Measure Integral and Real Analysis (2)

1. \((A \times C) / (B \times D) = [A \times (C / D)] \cup [(A / B) \times C]\)
2. \((A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D)\)
3. \((A \cup B) \times (C \cup D) \neq (A \times C) \cup (B \times D)\)

Integration

Integration w.r.t a Measure

Definition 3.1: \(S\)-Partition

Suppose \(S\) is a \(\sigma\)-algebra on a set \(X\). An \(S\)-partition of \(X\) is finite collection \(A_1, ..., A_m\) of disjoint sets in \(S\) s.t \(A_1 \cup ... \cup A_m = X\).

Definition 3.1: Lower Lebesgue Sum

Suppose \((X, S, \mu)\) is a measure space, \(f: X \rightarrow [0, \infty]\) is an \(S\)-measurable function, and \(P\) is an \(S\)-partition \(A_1, ..., A_m\) of \(X\). The lower Lebesgue sum \(L(f, P)\) is defined by:

\[L(f, P) = \sum^m_{j=1} \mu(A_j) \inf_{A_j} f\]

The definition does not assume \(X\) is a subset of \(\mathbb{R}\).

Definition 3.3: Integral of a Nonnegative Function

Suppose \((X, S, \mu)\) is a measure space and \(f: X \rightarrow [0, \infty]\) is an \(S\)-measurable function. The integral of \(f\) w.r.t \(\mu\), denoted by \(\int f d\mu\) is defined by:

\[\int f d\mu = \sup\{L(f, P): \text{ $P$ is an $S$-partition of $X$}\}\]

Theorem 3.4: Integral of a Characteristic Function \(\chi_E\)

Suppose \((X, S, \mu)\) is a measure space and \(E \in S\). Then:

\[\int \chi_E d\mu = \mu(E)\]

Proof of Theorem 3.4:

If \(P\) is the \(S\)-partition of \(E, X / E\), then:

\[L(\chi_E, P) = \mu(E) * 1 + 0 * \mu(X / E) = \mu(E)\]

Thus,

\[\int \chi_E d\mu \geq \mu(E)\]

Let \(P\) be an \(S\)-partition \(A_1, ..., A_m\) of \(X\), then

\[ \mu(A_j) \inf_{A_j} \chi_ E= \begin{cases} \mu(A_j), \quad \text{if } A_j \subseteq E\\ 0, \quad \text{ o. w } \end{cases} \]

Then:

\[\sum^{m}_{j=1} \mu(A_j) \inf_{A_j} \chi_E = \sum_{\{j: A_j \subseteq E\}} \mu(A_j) = \mu(\bigcup_{\{j: A_j \subseteq E\}} A_j) \leq \mu(E)\]

Theorem 3.7: Integral of a Simple Function

Suppose \((X, S, \mu)\) is a measure space, \(E_1, E_2, .., E_n\) are disjoint sets in \(S\), and \(c_1, ..., c_n \in [0, \infty]\). Then:

\[\int (\sum^n_{k=1} c_k \chi_{E_k}) d\mu = \sum^{n}_{k=1} c_k \mu(E_k)\]

Theorem 3.8: Integration is Order Preserving

Suppose \((X, S, \mu)\) is a measure space and \(f, g: X \rightarrow [0, \infty]\) are \(S\)-measurable functions s.t \(f(x) \leq g(x)\) for all \(x \in X\). Then:

\[\int f d\mu \leq \int g d\mu\]

Proof of Theorem 3.8:

Suppose \(P\) is a \(S\)-partition \(A_1, ..., A_n\) of \(X\). Then:

\[\inf_{A_i} f \leq \inf_{A_i} g \implies L(f, P) \leq L(g, P)\]

Then for any \(P\), we have:

\[\sup_P L(f, P) \leq \sup_P L(g, P)\]

Theorem 3.9: Integral Via Simple Functions

Suppose \((X, S, \mu)\) is a measure space and \(f: X \rightarrow [0, \infty]\) is \(S\)-measurable. Then:

\[\int f d\mu = \sup\{\sum^{m}_{j=1}c_j\mu(A_j): \text{ $A_1, ..., A_m$ are disjoint sets in $S$, $c_1, ...., c_m \in [0, \infty)$, $f(x)\geq \sum^m_{j=1}c_j \chi_{A_j}(x)$ for every $x \in X$}\}\]

Here, we have \(A_1, ..., A_m \in S\) as arbitrary disjoint sets, they do not have to be \(S\)-partition.

Theorem 3.11: Monotone Convergence Theorem

Suppose \((X, S, \mu)\) is a measure space and \(0 \leq f_1 \leq f_2 \leq ...\) is an increasing sequence of \(S\)-measurable functions. Define \(f: X \rightarrow [0, \infty]\) by:

\[f(x) = \lim_{k \rightarrow \infty} f_k (x)\]

Then:

\[\lim_{k \rightarrow \infty} \int f_k d\mu = \int f d\mu\]

If:

\[f(x) = \lim_{k \rightarrow \infty} \sum^{k}_{n=1}f_n (x)\]

Then:

\[\sum^{\infty}_{n=1} \int f_n d\mu = \int f d\mu\]

Proof of Theorem 3.11:

To show that the equality holds, we need to show that:

1. \(\lim_{k\rightarrow \infty} \int f_k d\mu \leq \int f d\mu\)
2. \(\lim_{k\rightarrow \infty} \int f_k d\mu \geq \int f d\mu\)

We first prove 1:

Suppose \((X, S, \mu)\) is a measure space and \(0 \leq f_1 \leq f_2 \leq ...\) is an increasing sequence of \(S\)-measurable functions. Define \(f: X \rightarrow [0, \infty]\) by:

\[f(x) = \lim_{k \rightarrow \infty} f_k (x)\]

Then by theorem 2.53, we have \(f\) to be \(S\)-measurable. Since \(f_k(x) \leq f(x), \;\; \forall x \in X, k \in \mathbb{Z}^+\), by theorem 3.8, we have:

\[\int f_k d \mu \leq \int f d\mu \implies \lim_{k \rightarrow \infty} \int f_k d \mu \leq \int f d\mu\]

To prove 2, we need to use theorem 3.9:

Suppose \(A_1, ..., A_m\) are disjoint sets in \(S\), and \(c_1, ..., c_m \in [0, \infty)\) are s.t:

\[f(x) \geq \sum^{m}_{j=1} c_j \chi_{A_j} (x)\]

Then:

\[\int f d\mu = \sup_{\text{all disjoint sets in $S$}}\{\sum^{m}_{j=1} c_j \mu(A_j)\}\]

Let \(t \in (0, 1)\). For \(k \in \mathbb{Z}^+\), let:

\[E_k = \{x \in X: f_k (x) \geq t \sum^{m}_{j=1} c_j \chi_{A_j} (x)\}\]

Then \(E_1 \subseteq E_2 \subseteq , ...\) is an increasing sequence of sets whose union is \(X\) (because \(f(x) \geq \sum^{m}_{j=1} c_j \chi_{A_j} (x)\)) and \((E_1 \cap A_j) \subseteq (E_2 \cap A_j) \subseteq .... \;\; \forall j \in \{1, ..., m\}\) is also an increasing sequence of sets, this implies (by theorem 2.59) that:

\[\mu(\lim_{k\rightarrow \infty} (E_k \cap A_j)) = \mu(X \cap A_j) = \mu(A_j)\]

Then for all \(x \in X\):

\[f_k (x) \geq t \sum^{m}_{j=1} c_j \chi_{A_j \cap E_k} (x) \implies \int f_k d\mu \geq \int t \sum^{m}_{j=1} c_j \chi_{A_j \cap E_k} d \mu \implies \int f_k d\mu \geq t\sum^{m}_{j=1} c_j \mu(A_j \cap E_k)\]

Taking the limit to infinity:

\[\lim_{k \rightarrow \infty} \int f_k d\mu \geq \lim_{k\rightarrow \infty} t\sum^{m}_{j=1} c_j \mu(A_j \cap E_k) = t \sum^{m}_{j=1} c_j \mu(A_j)\]

Taking the limit \(t \rightarrow 1\):

\[\lim_{k \rightarrow \infty} \int f_k d\mu \geq \sum^{m}_{j=1} c_j \mu(A_j)\]

Since right-hand side works for any \(c_1, ...., c_m \in [0, \infty)\), disjoint sets \(A_1, ..., A_m\) that satisfies \(f(x) \geq \sum^{m}_{j=1} c_j \chi_{A_j} (x)\), we have:

\[\lim_{k \rightarrow \infty} f_k(x) \geq \sup_{\text{all disjoint sets in $S$}}\{\sum^{m}_{j=1} c_j \mu(A_j)\} = \int f d\mu\]

Theorem 3.13: Integral-type Sums for Simple Functions

Suppose \((X, S, \mu)\) is a measurable space. Suppose \(a_1, ..., a_m, b_1, ..., b_n \in [0, \infty]\) and \(A_1, ..., A_m, B_1, ..., B_n \in S\) s.t \(\sum^m_{j=1} a_j \chi_{A_j} = \sum^n_{k=1} b_k \chi_{B_k}\). Then:

\[\sum^{m}_{j=1} a_j \mu(A_j) = \sum^n_{k=1} b_k \mu(B_k)\]

Theorem 3.15: Integral of a Linear Combination of Characteristic Functions

Suppose \((X, S, \mu)\) is a measure space, \(E_1, ..., E_n \in S\), and \(c_1, ..., c_n \in [0, \infty)\). Then:

\[\int (\sum^{n}_{k=1} c_k \chi_{E_k}) d\mu = \sum^{n}_{k=1} c_k \mu(E_k)\]

The difference between this theorem and theorem 3.7 is that, here we do not require \(E_1, ..., E_n\) to be disjoint

Theorem 3.16: Additivity of Integration

Suppose \((X . S, \mu)\) is a measure space and \(f, g: X \rightarrow [0, \infty]\) are \(S\)-measurable functions. Then:

\[\int (f+g) d\mu = \int f d\mu + \int g d\mu\]

Proof of Theorem 3.16:

Suppose \(f, g\) are simple functions, then:

\[\int (f + g) d\mu = \int (\sum_j c_j \chi_{E_j} + \sum_k c_k \chi_{E_k}) d\mu = \sum_j c_j \mu(E_j) + \sum_{k} c_k \mu(E_k) = \int f d\mu + \int g d\mu\]

So the integrals of simple functions are additive.

Let \(f_1, f_2, ...\), \(g_1, g_2, ....\) be increasing sequence of simple functions s.t:

\[\lim_{k\rightarrow \infty} f_k = f\]

\[\lim_{k\rightarrow \infty} g_k = g\]

By theorem 2.89, these sequences exist.

Since \(f_k, g_k, f, g\) are \(S\)-measurable functions, \(f + g\) is \(S\)-measurable function. Then by Monotone Convergence Theorem, we have:

\[\int (f + g) d\mu = \lim_{k \rightarrow \infty} \int (f_k + g_k) d\mu = \lim_{k \rightarrow \infty} \int f_k \mu + \lim_{k \rightarrow \infty} \int g_k d\mu = \int f d\mu + \int g d\mu\]

Definition 3.17: \(f^+, f^-\)

Suppose \(f: X \rightarrow [-\infty, \infty]\) is a function. Define function \(f^+\) and \(f^-\) from \(X\) to \([0, \infty]\) by:

\[ f^{+} (x) = \begin{cases} f^{+}(x), \quad \text{if } f(x) \geq 0\\ 0, \quad \text{if } f(x) < 0 \end{cases} \]

and

\[ f^{-} (x) = \begin{cases} 0, \quad \text{if } f(x) \geq 0\\ -f^{-}(x), \quad \text{if } f(x) < 0 \end{cases} \]

Then:

\[f = f^+ - f^-\]

and

\[|f| = f^+ + f^-\]

This decomposition allows us to extend our definition of integration to functions that take on negative as well as positive values.

Definition 3.18: Integral of a Real-Valued Function: \(\int f d\mu\)

Suppose \((X, S, \mu)\) is a measure space and \(f: X \rightarrow [-\infty, \infty]\) is an \(S\)-measurable function s.t at least one of \(\int f^+ d\mu\) and \(\int f^{-} d\mu\) is finite. The integral of \(f\) w.r.t \(\mu\), denoted \(\int f d\mu\) is defined by:

\[\int f d\mu = \int f^+ d\mu - \int f^- d\mu\]

The condition \(\int |f| d\mu < \infty\) is equivalent to the condition \(\int f^+ d\mu < \infty\) and \(\int f^- d\mu < \infty\)

Example 3.19

Suppose \(\lambda\) is a Lebesgue measure on \(\mathbb{R}\) and \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a function defined by:

\[ f (x) = \begin{cases} 1, \quad \text{if } x \geq 0\\ 0, \quad \text{if } x < 0 \end{cases} \] Then \(\int f^+ d\lambda = \infty\) and \(\int f^- d\lambda = \infty\), so \(\int f d\lambda\) is not defined.

Theorem 3.20: Integration is Homogeneous

Suppose \((X, S, \mu)\) is a measure space and \(f: X \rightarrow [-\infty, \infty]\) is a function such that \(\int f d\mu\) is defined. If \(c \in \mathbb{R}\), then:

\[\int cf d\mu = c\int f d\mu\]

Theorem 3.21: Additivity of Integration

Suppose \((X, S, \mu)\) is a measure space and \(f, g: X \rightarrow \mathbb{R}\) are \(S\)-measurable functions s.t \(\int |f| d\mu < \infty\) and \(\int |g| d\mu < \infty\). Then:

\[\int (f + g) d\mu = \int f d\mu + \int g d\mu\]

Theorem 3.22: Integration is Order Preserving

Suppose \((X, S ,\mu)\) is a measure space and \(f, g: X \rightarrow \mathbb{R}\) are \(S\)-measurable functions s.t \(\int f d\mu\) and \(\int g d\mu\) are defined. Suppose also that \(f(x) \leq g(x)\) for all \(x \in X\). Then \(\int f d\mu \leq \int g d\mu\).

Theorem 3.23: Absolute Value of Integral \(\leq\) Integral of Absolute Value

Suppose \((X, S, \mu)\) is a measure space and \(f: X \rightarrow [-\infty, \infty]\) is a function s.t \(\int f d\mu\) is defined. Then:

\[|\int f d\mu| \leq \int |f| d\mu\]

Limits of Integrals and Integrals of Limits

Definition 3.24: Integration on a Subset; \(\int_E f d\mu\)

Suppose \((X, S, \mu)\) is a measure space and \(E \in S\). If \(f: X \rightarrow [-\infty, \infty]\) is an \(S\)-measurable function, then \(\int_E f d\mu\) is defined by:

\[\int_E f d\mu = \int \chi_{E} f d\mu\]

If the right side of the equation above is defined, o.w \(\int_E f d\mu\) is undefined.

We can also think of \(\int_E f d\mu\) as \(\int f_E d\mu_E\) where \(\mu_E\) is the measure botained by restricting \(\mu\) to the elements of \(S\) that are contained in \(E\).

Notice that, \(\int_X f d\mu\) means the same as \(\int f d\mu\).

Theorem 3.25: Bounded an Integral

Suppose \((X, S, \mu)\) is a measure space, \(E \in S\) and \(f: X \rightarrow [-\infty, \infty]\) is a function s.t \(\int_E fd\mu\) is defined. Then:

\[| \int_E fd\mu| \leq \mu(E) \sup_E |f|\]

Proof Theorem 3.25:

Let \(c = \sup_E |f|\):

\[|\int_E f d\mu| = |\int \chi_E f d\mu| \leq \int \chi_E |f| d\mu \leq \int \chi_E c d\mu = c\mu(E)\]

Theorem 3.26: Bounded Convergence Theorem

Suppose \((X, S, \mu)\) is a measure space and \(\mu(X) < \infty\). Suppose \(f_1, f_2, ...\) is a sequence of \(S\)-measurable functions from \(X \rightarrow \mathbb{R}\). If there exists \(c \in (0, \infty)\) s.t:

\[|f_k (x) \leq c|\]

for all \(k \in \mathbb{Z}^+\) and all \(x \in X\), then:

\[\lim_{k \rightarrow \infty} \int f_k d\mu = \int f d\mu\]

Proof of Theorem 3.26:

The function \(f\) is \(S\)-measurable by theorem 2.48. Suppose \(c\) satisfies the assumption above. Let \(\epsilon > 0\), by Theorem 2.85, there exists \(E \in S\), s.t \(\mu(S/ E) < \frac{\epsilon}{4c}\) and \(f_1, f_2, ...\) converges uniformly to \(f\) on \(E\). Now:

\[|\int f_k d\mu - \int f d\mu| = |\int_{X / E} (f_k - f) d\mu + \int_E (f_k - f)d\mu| = |\int_{X / E} f_k d\mu - \int_{X / E} f d\mu + \int_E (f_k - f)d\mu|\]

Then by theorem 3.23:

\[|\int_{X / E} f_k d\mu - \int_{X / E} f d\mu + \int_E (f_k - f)d\mu| \leq \int_{X / E} |f_k| d\mu + \int_{X / E} |f| d\mu + \int_E |f_k - f|d\mu|\]

Since \(|f_k (x)| \leq c \forall x \in X\) and by theorem 3.25:

\[\int_{X / E} |f_k| d\mu + \int_{X / E} |f| d\mu + \int_E |f_k - f|d\mu| \leq \int c\chi_{X/E} d\mu + \int c\chi_{X/E} d\mu + \int_E |f_k - f|d\mu| \leq \frac{\epsilon}{2} + \mu(E)\sup_E |f_k - f|\]

Since \(f_k\) converges uniformly to \(f\) on \(E\), for \(k\) sufficient large, we have \(|f_k(x) - f(x)| < \epsilon, \;\forall x\in X\), this implies that:

\[\lim_{k\rightarrow \infty} |\int f_k d\mu - \int f d\mu| = 0\]

Definition 3.27: Almost Every

Suppose \((X, S, \mu)\) is a measure space. A set \(E \in S\) is said to contain \(\mu-almost\) every element of \(X\) if \(\mu(X / E) = 0\). If the measure \(\mu\) is clear from the context, then the phrase almost every can be used.

For example, almost every real number is irrational w.r.t the usual Lebesgue measure because \(|\mathbb{Q}| = 0\).

Theorems about integrals can almost always be relaxed so that the hypotheses apply only almost everywhere instead of everywhere.

Theorem 3.28: Integral on Small Sets are Small

Suppose \((X, S, \mu)\) is a measure space, \(g: X \rightarrow [0, \infty]\) is \(S\)-measurable, and \(\int g d\mu < \infty\). Then for every \(\epsilon > 0\), there exists \(\delta > 0\) s.t:

\[\int_B g d\mu < \epsilon\]

For every set \(B \in S\) s.t \(\mu(B) < \delta\).

Theorem 3.29: Integrable Functions Live Mostly on Sets of Finite Measure

Suppose \((X, S, \mu)\) is a measure space, \(g: X \rightarrow [0, \infty]\) is \(S\)-measurable, and \(\int g d\mu < \infty\). Then for every \(\epsilon > 0\), there exists \(E \in S\) s.t \(\mu(E) < \infty\) and

\[\int_{X/ E} g d\mu < \epsilon\]

Theorem 3.31: Dominated Convergence Theorem

Suppose \((X, S, \mu)\) is a measure space, \(f: X \rightarrow [-\infty, \infty]\) is \(S\)-measurable, and \(f_1, f_2, ...\) are \(S\)-measurable functions from \(X\) to \([-\infty, \infty]\) s.t:

\[\lim_{k\rightarrow \infty} f_k(x) = f(x)\]

for almost every \(x \in X\). If there exists an \(S\)-measurable function \(g: X \rightarrow [0, \infty]\) s.t:

\[\int gd\mu < \infty\]

and

\[|f_k(x) | \leq g(x)\]

for every \(k \in \mathbb{Z}^+\) and almost every \(x \in X\), then:

\[\lim_{k \rightarrow \infty} \int f_k d\mu = \int f d\mu\]

Theorem 3.34: Riemann Integrable \(\Longleftrightarrow\) Continuous Almost Everywhere

Suppose \(a < b\) and \(f: [a, b] \rightarrow \mathbb{R}\) is a bounded function. Then \(f\) is Riemann integrable IFF:

\[|\{x \in [a, b]: \text{ $f$ is not continuous at $x$}\}| = 0\]

This implies that the outer measure of the set of discontinuities equals to 0 (The set of discontinuities is countable).

Furthermore, if \(f\) is Riemann integrable and \(\lambda\) denotes Lebesgue measure on \(\mathbb{R}\), then \(f\) is Lebesgue measurable and:

\[\int^b_a f = \int_{[a, b]} f d\lambda\]

Definition 3.39: \(\int^b_a f\)

We previously defined the notation \(\int^b_a f\) to mean the Riemann integral of \(f\), now we redefine \(\int^b_a f\) to denote the Lebesgue integral.

Suppose \(-\infty \leq a < b \leq \infty\) and \(f: (a, b) \rightarrow \mathbb{R}\) is Lebesgue measurable. Then:

• \(\int^b_a f\) and \(\int^b_a f(x) dx\) mean \(\int_{(a, b)} f d\lambda\), where \(\lambda\) is Lebesgue measure on \(\mathbb{R}\).
• \(\int^b_a f\) is defined to be \(- \int^b_a f\).

Definition 3.40: \(\|f\|; L^1(\mu)\)

Suppose \((X, S, \mu)\) is a measure space. If \(f: X \rightarrow [-\infty, \infty]\) is \(S\)-measurable, then the \(L^1\)-norm of \(f\) is denoted by \(\|f\|_1\) and is defined by:

\[\|f\|_1 = \int |f| d\mu\]

The Lebesgue space \(L^1(\mu)\) is defined by:

\[L^1(\mu) = \{f: f \text{ $f$ is an $S$-measurable function from $X \rightarrow \mathbb{R}$ and $\|f\|_1 < \infty$}\}\]

Example 3.41:

Suppose \((X, S, \mu)\) is a measure space and \(E_1, ..., E_n\) are disjoint subsets of \(X\). Suppose \(a_1, ..., a_n\) are distinct nonzero real numbers. Then:

\[a_1 \chi_{E_1} + ... + a_n \chi_{E_n} \in L^1(\mu)\]

IFF \(E_k \in S\) and \(\mu(E_k) < \infty\).

Theorem 3.43: Properties of the \(L^1-\)norm

Suppose \((X, S, \mu)\) is a measure space and \(f, g \in L^1(\mu)\), then:

• \(\|f\|_1 \geq 0\)
• \(\|f\|_1 = 0\) IFF \(f(x) = 0\) for almost every \(x \in X\)
• \(\|cf\|_1 = |c|\|f\|_1, \; \forall c \in \mathbb{R}\)
• \(\|f+g\|_1 \leq \|f\|_1 + \|g\|_1\)

Theorem 3.44: Approximation by Simple Functions

Suppose \(\mu\) is a measure and \(f \in L^1 (\mu)\). Then for every \(\epsilon > 0\), there exists a simple function \(g \in L^1(\mu)\) s.t:

\[\|f - g\|_1 < \epsilon\]

Definition 3.45: \(L^1(\mathbb{R}); \|f\|_1\)

• The notation \(L^1(\mathbb{R})\) denotes \(L^1(\lambda)\), where \(\lambda\) is Lebesgue measure on either the Borel subsets of \(\mathbb{R}\) or the Lebesgue measurable subsets of \(\mathbb{R}\).
• When working with \(L^1 (\mathbb{R})\), the notation \(\|f\|_1\) denotes the integral of the absolute value of \(f\) w.r.t Lebesgue measure on \(\mathbb{R}\).

Definition 3.46: Step Function

A step function is a function \(g: \mathbb{R} \rightarrow \mathbb{R}\) of the form:

\[g = a_1 \chi_{I_1} + .... + a_n \chi_{I_n}\]

Where \(I_1, ...., I_n\) are intervals of \(\mathbb{R}\) and \(a_1, ..., a_n\) are nonzero real numbers. The intervals can be open, closed or half-open intervals.

Compare with simple function, the sets \(I_1, ..., I_n\) are required to be intervals in step functions.

Theorem 3.47: Approximation by Step Functions

Suppose \(f \in L^1 (\mathbb{R})\). Then for every \(\epsilon > 0\), there exists a step function \(g \in L^1 (\mathbb{R})\) s.t:

\[||f - g\|_1 < \epsilon\]

Theorem 3.48: Approximation by Continuous Functions

Suppose \(f \in L^1(\mathbb{R})\). Then for every \(\epsilon > 0\), there exists a continuous function \(g: \mathbb{R} \rightarrow \mathbb{R}\) s.t:

\[\|f - g\|_1 < \epsilon\]

and \(\{x \in \mathbb{R}: g(x) \neq 0\}\) is a bounded set.

Differentiation

Hardy-Littlewood Maximal Function

Theorem 4.1: Markov's Inequality

Suppose \((X, S, \mu)\) is a measure space and \(h \in L^1 (\mu)\). Then:

\[\mu(\{x \in X: |h(x)| \geq c\}) \leq \frac{1}{c} \|h\|_1\]

for every \(c > 0\).

Proof of Theorem 4.1:

Suppose \(c > 0\). Then:

\[\begin{aligned} \mu(\{x \in X: |h(x)| \geq c\}) &= \int_{\{x \in X: |h(x)| \geq c\}} 1 d\mu \\ &= \frac{1}{c}\int_{\{x \in X: |h(x)| \geq c\}} c d\mu\\ &\leq \frac{1}{c}\int_{\{x \in X: |h(x)| \geq c\}} |h| d\mu\\ &\leq \frac{1}{c}\int_{\{x \in X: |h(x)| \geq c\}} |h| d\mu + \frac{1}{c}\int_{\{x \in X: |h(x)| < c\}} |h| d\mu\\ &=\frac{1}{c}\|h\|_1 \end{aligned}\]

Definition 4.2: \(3\) times a Bounded Nonempty Open Interval

Suppose \(I\) is a bounded nonempty open interval of \(\mathbb{R}\). Then \(3 * I\) denotes the open interval with the same center as \(I\) and three times the length of \(I\).

If \(I = (0, 10)\), \(3 * I = (-10, 20)\)

Theorem 4.4: Vitali Covering Lemma

Suppose \(I_1, ..., I_n\) is a list of bounded nonempty open intervals of \(\mathbb{R}\). Then there exists a disjoint sublist \(I_{k_1}, ..., I_{k_m}\) s.t:

\[I_1 \cup ... \cup I_n \subseteq (3 * I_{k_1}) \cup .... \cup (3 * I_{k_n})\]

Example 4.5:

Suppose \(n = 4\) and

\[I_1 = (0, 10), I_2 = (9, 15), I_3 = (14, 22), I_4 = (21, 31)\]

Then:

\[3 * I_1 = (-10, 20), 3 * I_2 = (3, 21), 3 * I_3 = (6, 30), 3 * I_4 = (11, 41)\]

Thus, \(I_1 \cup I_2 \cup I_3 \cup I_4 \subseteq (3 * I_1) \cup (3 * I_4)\)

In this example, \(I_1, I_4\) is the only sublist of \(I_1, ..., I_4\) that produces the conclusion of Vitali Covering Lemma.

Definition 4.6: Hardy-Littlewood Maximal Function; \(h^*\)

Suppose \(h: \mathbb{R} \rightarrow \mathbb{R}\) is a Lebesgue measurable function. Then the Hardy-Littlewood maximal function of \(h\) is the function \(h^* : \mathbb{R} \rightarrow [0, \infty]\) defined by:

\[h^*(b) = \sup_{t > 0} \frac{1}{2t}\int^{b+t}_{b-t} |h| = \sup_{t > 0} \frac{1}{2t}\int_{(b+t, b-t)} |h| d\lambda\]

Where \(h^*\) is a Borel measurable function.

In other words, \(h^*(b)\) is the supremum over all bounded intervals centered at \(b\) of average of \(|b|\) on those intervals.

Theorem 4.8: Hardy-Littlewood Maximal Inequality

Suppose \(h \in L^1 (\mathbb{R})\). Then:

\[|\{b \in \mathbb{R}: h^*(b) > c\}| \leq \frac{3}{c}\|h\|_1\]

Derivatives of Integrals

Theorem 4.10: Lebesgue Differentiation Theorem, First Version

Suppose \(f \in L^1(\mathbb{R})\). Then:

\[\lim_{t \rightarrow 0} \frac{1}{2t} \int^{b+t}_{b-t} |f - f(b)| = 0\]

for almost every \(b \in \mathbb{R}\)

In other words, the average amount by which a function in \(L^1(\mathbb{R})\) differs from its values is small almost everywhere on small intervals.

Definition 4.16: Derivative; \(g^{\prime}\); Differentiable

Suppose \(g: I \rightarrow \mathbb{R}\) is a function defined on an open interval \(I\) of \(\mathbb{R}\) and \(b \in I\). Then derivative of \(g\) at \(b\), denoted \(g^{\prime}(b)\) is defined by:

\[g^{\prime}(b) = \lim_{t \rightarrow 0} \frac{g(b + t) - g(b)}{t}\]

If the limit above exists, in which case \(g\) is called differentiable at \(b\).

Theorem 4.17: Fundamental Theorem of Calculus

Suppose \(f \in L^1(\mathbb{R})\). Define \(g: \mathbb{R} \rightarrow \mathbb{R}\) by:

\[g(x) = \int^x_{-\infty} f\]

Suppose \(b \in \mathbb{R}\) and \(f\) is continuous at \(b\). Then \(g\) is differentiable at \(b\) and:

\[g^{\prime}(b) = f(b)\]

Proof of Theorem 4.17:

If \(t \neq 0\), then:

\[\begin{aligned} |\frac{g(b + t) - g(b)}{t} - f(b)| &= |\frac{\int^{b+t}_{-\infty} f - \int^b_{-\infty} f}{t} - f(b)|\\ &=|\frac{\int^{b+t}_{b} f}{t} - f(b)|\\ &=|\frac{\int^{b+t}_{b} f}{t} - \frac{\int^{b+t}_{b} f(b)}{t}|\\ &=|\frac{\int^{b+t}_{b} f - f(b)}{t}|\\ &\leq \sup_{\{x \in \mathbb{R}: |x - b| < |t|\}} |f(x) - f(b)| \end{aligned}\]

Since \(f\) is continuous at \(b\), for all \(\epsilon > 0\), there exists \(\delta > 0\), s.t \(|x - b| < \delta \implies |f(x) - f(b)| < \epsilon\), if we let \(|t| \rightarrow 0\), s.t \(|t| < \delta, \; \forall \delta > 0\), then:

\[\lim_{t \rightarrow 0} \sup_{\{x \in \mathbb{R}: |x - b| < |t|\}} |f(x) - f(b)| < \epsilon\]

Thus, \(g^{\prime}(b) = f(b)\).

Theorem 4.19: Lebesgue Differentiation Theorem, Second Version

Suppose \(f \in L^{1} (\mathbb{R})\). Define \(g: \mathbb{R} \rightarrow \mathbb{R}\) by:

\[g(x) = \int^x_{-\infty} f\]

Then \(g^{\prime} (b) = f(b)\) for almost every \(b \in \mathbb{R}\)

Theorem 4.21: \(L^1 (\mathbb{R})\) Function Equals Its Local Average Almost Everywhere

Suppose \(f \in L^1 (\mathbb{R})\). Then:

\[f(b) = \lim_{t \rightarrow 0} \frac{1}{2t} \int^{b+t}_{b-t} f\]

for almost every \(b \in \mathbb{R}\).

The results holds for at every number \(b\) at which \(f\) is continuous. Even if \(f\) is discontinuous everywhere, the conclusion holds for almost every real number \(b\).

Definition 4.22: Density

Suppose \(E \subseteq \mathbb{R}\). The density of \(E\) at a number \(b \in \mathbb{R}\) is:

\[\lim_{t \rightarrow 0} \frac{|E \cap (b-t, b+t)|}{2t}\]

If this limit exists (o.w the density of \(E\) at \(b\) is undefined).

Example 4.23

\[ \text{The density of $[0, 1]$ at $b$ } = \begin{cases} 1, \quad &\text{If } b \in (0, 1)\\ \frac{1}{2}, \quad &\text{If $b = 0$ or $b = 1$}\\ 0, \quad & o.w \end{cases} \]

Theorem 4.24: Lebesgue Density Theorem

Suppose \(E \subseteq \mathbb{R}\) is a Lebesgue measurable set. Then the density of \(E\) is \(1\) at almost every elment of \(E\) and is \(0\) at almost every element of \(\mathbb{R} / E\).

Theorem 4.25: Bad Borel Set

There exists a Borel set \(E \subseteq \mathbb{R}\) s.t

\[0 < |E \cap I| < |I|\]

For every nonempty bounded open interval \(I\).

Product Measures

Products of Measures

Definition 5.1: Rectangle

Suppose \(X, Y\) are sets. A rectangle in \(X \times Y\) is a set of the form \(A \times B\), where \(A \subseteq X, B \subseteq Y\).

Definition 5.2: Product of Two \(\sigma\)-algebras; \(S \otimes T\); Measurable Rectangle

Suppose \((X, S)\) and \((Y, T)\) are measurable spaces. Then:

• The product \(S \otimes T\) is defined to be the smallest \(\sigma\)-algebra on \(X \times Y\) that contains: \[\{A \times B: A \in S, B \in T\}\]
• A measurable rectangle in \(S \otimes T\) is a set of the form \(A \times B\), where \(A \in S\) and \(B \in T\).

Thus, \(S \otimes T\) is the smallest \(\sigma\)-algebra that contains all the measurable rectangles in \(S \otimes T\).

Definition 5.3: Cross Sections of Sets; \(E_a\) and \(E^b\)

Suppose \(X, Y\) are sets and \(E \subseteq X \times Y\). Then for \(a \in X\) and \(b \in Y\), the cross sections \(E_a\) and \(E^b\) are defined by:

\[E_a = \{y \in Y: (a, y) \in E\}, \quad E^b = \{x \in X: (x, b) \in E\}\]

Example 5.5:

Suppose \(X, Y\) are sets and \(A \subseteq X, B \subseteq Y\), then:

\[ [A \times B]_a = \begin{cases} B, \quad & \text{if } a \in A\\ \emptyset, \quad & \text{if } a \notin A \end{cases} \]

and

\[ [A \times B]^b = \begin{cases} A, \quad & \text{if } b \in B\\ \emptyset, \quad & \text{if } b \notin B \end{cases} \]

Theorem 5.6: Cross Sections of Measurable sets are Measurable

Suppose \(S\) is a \(\sigma\)-algebra on \(X\) and \(T\) is \(\sigma\)-algebra on \(Y\). If \(E \in S \otimes T\), then

\[E_a \in T, \; \forall a \in X , \quad\quad E^b \in S, \; \forall b \in Y\]

In other words, Let \(\varepsilon\) denotes the collection of subsets \(E\) of \(X \times Y\) for which the conclusion of this result holds. Then \(\varepsilon\) is \(\sigma\)-algebra that contains \(S \otimes T\).

Definition 5.7: Cross Sections of Functions; \(f_a, f^b\)

Suppose \(X, Y\) are sets and \(f: X \times Y \rightarrow \mathbb{R}\) is a function. Then for \(a \in X\) and \(b \in Y\), the cross section functions \(f_a: Y \rightarrow \mathbb{R}\) and \(f^b: X \rightarrow \mathbb{R}\) are defined by:

\[f_a (y) = f(a, y), \quad \forall y \in Y\]

and

\[f^b (x) = f(x , b), \quad \forall x \in X\]

Theorem 5.9: Cross Sections of Measurable Functions are Measurable

Suppose \(S\) is a \(\sigma\)-algebra on \(X\) and \(T\) is a \(\sigma\)-algebra on \(X\) and \(T\) is a \(\sigma\)-algebra on \(Y\). Suppose \(f: X \times Y \rightarrow \mathbb{R}\) is an \(S \otimes T\)-measurable function. Then:

\[f_a \text{ is a $T$-measurable function on $Y$ for every $a \in X$}\]

and

\[f^b \text{ is a $S$-measurable function on $X$ for every $b \in Y$}\]

Proof of Theorem 5.9:

Suppose \(D\) is a Borel set of \(\mathbb{R}\) and \(a \in X\). For all \(y \in Y\), we want to show that:

\[y \in (f_a)^{-1}(D) \Longleftrightarrow y \in [f^{-1}(D)]_a\]

Then,

\[\begin{aligned} y \in (f_a)^{-1}(D) &\Longleftrightarrow f(a, y) \in D\\ &\Longleftrightarrow (a, y) \in f^{-1} (D)\\ &\Longleftrightarrow y \in [f^{-1}(D)]_a\\ \end{aligned}\]

Thus, \((f_a)^{-1}(D) = [f^{-1}(D)]_a\)

If \(f\) is \(S \otimes T\)-measurable, then:

\[f^{-1}(D) \in S \otimes T \implies [f^{-1}(D)]_a \in T \implies (f_a)^{-1}(D) \in T \implies (f_a)^{-1} \text{ is $T$-measurable}\]

Monotone Class Theorem

The following standard two-step technique often works to prove that every set in a \(\sigma\)-algebra has a certain property:

1. Show that Every set in a collection of sets that generates the \(\sigma\)-algebra has the property.
2. Show that the collection of sets that has the property is a \(\sigma\)-algebra.

Definition 5.10: Algebra

Suppose \(W\) is set and \(A\) is a set of subsets of \(W\). Then \(A\) is called an algebra on \(W\) if the following three conditions are satisfied:

1. \(\emptyset \in A\).
2. If \(E \in A\), then \(W / E \in A\).
3. If \(E, F \in A\), then \(E \cup F \in A\).

Thus, an algebra is closed under complementation and under finite unions. A \(\sigma\)-algebra is closed under complementation and countable unions.

Theorem 5.13: The Set of Finite Unions of Measurable Rectangles is an Algebra

Suppose \((X, S), (Y, T)\) are measurable spaces. Then:

1. The set of finite unions of measurable rectangles in \(S \otimes T\) is an algebra on \(X \times Y\).

2. Every finite union of measurable rectangles in \(S \otimes T\) can be written as a finite union of disjoint measurable rectangles in \(S \otimes T\).

   \[(A \times B) \cup (C \times D) = (A \times B) \cup (C \times (D / B)) \cup ((C / A) \times (B \cap D))\]

Definition 5.15: Monotone Class

Suppose \(W\) is a set and \(M\) is a set of subsets of \(W\). Then \(M\) is called a Monotone class on \(W\) if the following two conditions are satisfied:

• If \(E_1 \subseteq E_2 \subseteq ...\) is an increasing sequence of sets in \(M\), then: \[\bigcup^{\infty}_{k=1} E_k \in M\]
• If \(E_1 \supseteq E_2 \supseteq ...\) is an decreasing sequence of sets in \(M\), then: \[\bigcap^{\infty}_{k=1} E_k \in M\]

Every \(\sigma\)-algebra is a monotone class. However, some monotone classes are not closed under even finite unions.

Theorem 5.17: Monotone Class Theorem

Suppose \(A\) is an algebra on a set \(W\). Then the smallest \(\sigma\)-algebra containing \(A\) is the smallest monotone class containing \(A\). (The smallest monotone class is the intersection of all monotone classes on \(W\) that contain \(A\).)

Definition 5.18: Finite Measure; \(\sigma\)-Finite Measure

• A measure \(\mu\) on a measurable space \((X, S)\) is called finite if \(\mu(X) < \infty\).
• A measure is called \(\sigma\)-finite if the whole space can be written as the countable union of sets with finite measure.
• More precisely, a measure \(\mu\) on a measurable space \((X, S)\) is called -finite if there exists a sequence \(X_1, X_2, ...\) of sets in \(S\) s.t: \[X = \bigcup^{\infty}_{k=1} X_k, \quad \quad \mu(X_k) < \infty, \quad \forall k \in \mathbb{Z}^+\]

Theorem 5.20: Measure of Cross Section is a Measurable Function

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are \(\sigma\)-finite measure spaces. If \(E \in S \otimes T\), then:

1. \(x \longmapsto v(E_x)\) is an \(S\)-measurable function on \(X\) (Think of the function as \(f(x) = v(E_x)\)).
2. \(y \longmapsto \mu(E^y)\) is a \(T\)-measurable function on \(Y\) (Think of the function as \(g(y) = \mu(E^y)\)).

Definition 5.21: Integration Notation

Suppose \((X, S, \mu)\) is a measure space and \(g: X \rightarrow [-\infty, \infty]\) is a function. The notation:

\[\int g(x) d\mu(x) := \int g d\mu\]

Where \(d\mu(x)\) indicates that variables other than \(x\) should be treated as constants.

Definition 5.23: Iterated Integrals

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are measure spaces and \(f: X \times Y \rightarrow \mathbb{R}\) is a function. Then:

\[\int_{X} \int_{Y} f(x, y) dv(y) d\mu(x)\]

means:

\[\int_x (\int_{Y} f(x, y) dv(y)) d\mu(x)\]

In other words, to compute the integral above, first fix \(x \in X\) and compute \(\int_{Y} f(x, y) dv(y)\) if this integral makes sense. Then compute the integral w.r.t \(\mu\) of the function:

\[x \longmapsto \int_Y f(x, y) dv(y)\]

If this integral makes sense.

Definition 5.25: Product of Two Measures; \(\mu \times v\)

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are \(\sigma\)-finite measure spaces. For \(E \in S \otimes T\), define \((\mu \times v) (E)\) by:

\[(\mu \times v) (E) = \int_X \int_Y \chi_{E} (x, y) dv(y) d\mu(x)\]

Example 5.26

Suppose \((X, S, \mu), (Y, T, v)\) are \(\sigma\)-finite measure spaces. If \(A \in S\) and \(B \in T\), then: \[(\mu \times v) (A \times B) = \int_{X} \int_{Y} \chi_{A \times B} (x, y) dv(y) d\mu(x) = \int_{X} \int_{Y} \chi_{B} (y) \chi_{A} (x) dv(y) d\mu(x) = \int_{X} \chi_{A} (x) v(B) d\mu(x) = \mu(A) v(B)\]

Product measure of a measurable rectangle is the product of the measures of the corresponding sets.

Theorem 5.27: Product of Two Measures is a Measure

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are \(\sigma\)-finite measure spaces. Then \(\mu \times v\) is a measure on \((X \times Y, S \otimes T)\).

Theorem 5.28: Tonelli's Theorem

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are \(\sigma\)-finite measure spaces. Suppose \(f: X \times Y \rightarrow [0, \infty]\) is \(S \otimes T\) measurable. Then:

\[x \longmapsto \int_Y f(x, y) dv(y) \text{ is an $S$-measurable function on $X$}\]

\[y \longmapsto \int_X f(x, y) d\mu(x) \text{ is an $T$-measurable function on $Y$}\]

and

\[\int_{X \times Y} f d(\mu \times v) = \int_X \int_Y f(x, y) dv(y) d\mu(x) = \int_Y \int_X f(x, y) d\mu(x) dv(y)\]

Theorem 5.31: Double Sums of Nonnegative Numbers

If \(x_{j, k}: j, k \in \mathbb{Z}^+\) is a doubly indexed collection of nonnegative numbers, then:

\[\sum^\infty_{j=1}\sum^{\infty}_{k=1} x_{j, k} = \sum^{\infty}_{k=1} \sum^{\infty}_{j=1} x_{j, k}\]

Theorem 5.32: Fubini's Theorem

Suppose \((X, S, \mu)\) and \((Y, T, v)\) are \(\sigma\)-finite measure spaces. Suppose \(f: X \times Y \rightarrow [-\infty, \infty]\) is \(S \otimes T\)-measurable and \(\int_{X \times Y} |f| d(\mu \times v) < \infty\), then:

\[\int_Y |f(x, y)| dv(y) < \infty \text{ for almost every $x \in X$}\]

and

\[\int_Y |f(x, y)| d\mu(x) < \infty \text{ for almost every $y \in Y$}\]

Furthermore:

\[x \longmapsto \int_Y f(x, y) dv(y) \text{ is an $S$-measurable function on $X$}\]

\[y \longmapsto \int_X f(x, y) d\mu(x) \text{ is an $T$-measurable function on $Y$}\]

And:

\[\int_{X \times Y} f d(\mu \times v) = \int_X \int_Y f(x, y) dv(y) d\mu(x) = \int_Y \int_X f(x, y) d\mu(x) dv(y)\]

Definition 5.34: Region Under the Graph; \(U_f\)

Suppose \(X\) is a set and \(f: X \rightarrow [0, \infty]\) is a function. Then the region under the graph of \(f\), denoted \(U_f\) is defined by:

\[U_f = \{(x, t) \in X \times (0, \infty): 0 < t < f(x)\}\]

Theorem 5.35: Area Under the Graph of a Function Equals the Integral

Suppose \((X, S, \mu)\) is a \(\sigma\)-finite measure space and \(f: X \rightarrow [0, \infty]\) is an \(S\)-measurable function. Let \(B\) denotes the \(\sigma\)-algebra subsets of \((0, \infty)\), and let \(\lambda\) denotes the Lebesgue measure on \(((0, \infty), B)\). Then \(U_f \in S \otimes B\) and:

\[(\mu \times \lambda) (U_f) = \int_X f d\mu = \int_{(0, \infty)} \mu(\{x \in X: t < f(x)\}) d\lambda(t)\]

Lebesgue Integration on \(\mathbb{R}^n\)

Assume throughout this section, \(m, n\) are positive integers.

1. An open cube \(B(x, \delta)\) with side length \(2\delta\) is defined by: \[B(x, \delta) = \{y \in \mathbb{R}^n: \|y - x\|_{\infty} < \delta\}\]
2. A subset \(G \in \mathbb{R}^n\) is called open if for every \(x \in G\), there exists \(\delta > 0\) s.t \(B(x, \delta) \subseteq G\).
3. The union of every collection (finite or infinite) of open subsets of \(\mathbb{R}^n\) is an open subset of \(\mathbb{R}\).
4. The intersection of every finite collection of open subsets of \(\mathbb{R}^n\) is an open subset of \(\mathbb{R}^n\).
5. A subset of \(\mathbb{R}^n\) is called closed if its complement in \(\mathbb{R}^n\) is open.
6. A set \(A \subseteq \mathbb{R}^n\) is called bounded if \(\sup\{\|a\|_{\infty}: a \in A\} < \infty\).
7. \(\mathbb{R}^m \times \mathbb{R}^n\) is identified with \(\mathbb{R}^{m + n}\), note that \(\mathbb{R}^2 \times \mathbb{R}^{1} \neq \mathbb{R}^3\) however we can identify \(((x_1, x_2), x_3)\) with \((x_1, x_2, x_3)\), so we say that they are equal.
8. \(B(x, \delta) \times B(y, \delta) = B((x, y), \delta) \in \mathbb{R}^{m + n}, x \in \mathbb{R}^m, y \in \mathbb{R}^n\)
   

Theorem 5.36: Product of Open Set is Open

Suppose \(G_1\) is an open subset of \(\mathbb{R}^m\) and \(G_2\) is an open subset of \(\mathbb{R}^n\). Then \(G_1 \times G_2\) is an open subset of \(\mathbb{R}^{m+n}\).

Definition 5.37: Borel set; \(B_n\)

• A Borel set of \(\mathbb{R}^n\) is an element of the smallest \(\sigma\)-algebra on \(\mathbb{R}^n\) that containing all open subsets of \(\mathbb{R}^n\).
• The \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}^n\) is denoted by \(B_n\).

Theorem 5.38: Open Sets are Countable Unions of Open Cubes

1. A subset of \(\mathbb{R}^n\) is open in \(\mathbb{R}^n\) if and only if it is a countable union of open cubes in \(\mathbb{R}^n\).
2. \(B_n\) is the smallest \(\sigma\)-algebra on \(\mathbb{R}^n\) containing all open cubes in \(\mathbb{R}^n\).

Theorem 5.39: Product of the Borel Subsets of \(\mathbb{R}^m\) and the Borel Subsets of \(\mathbb{R}^n\)

\[B_{m} \otimes B_{n} = B_{m + n}\]

In other words, \(B_{m + n}\) is the smallest \(\sigma\)-algebra on \(\mathbb{R}^{m + n}\) that containing all measurable rectangles \(\{A \times B: A \subseteq B_m, B \subseteq B_n\}\).

And we can define \(B_m \otimes B_n \otimes B_p\) directly as the smallest \(\sigma\)-algebra on \(\mathbb{R}^{m + n + p}\) that containing \(\{A \times B \times C: A \subseteq B_m, B \subseteq B_n, C \subseteq B_p\}\)

Definition 5.40: Lebesgue Measure; \(\lambda_n\)

Lebesgue measure on \(\mathbb{R}^n\) is denoted by \(\lambda_n\) and is defined indutively by:

\[\lambda_n = \lambda_{n-1} \times \lambda_1\]

where \(\lambda_1\) is the Lebesgue measure on \((\mathbb{R}, B_1)\)

or identically, we can write:

\[\lambda_n = \lambda_{j} \times \lambda_{k}\]

where \(j, k \in \mathbb{Z}^+, j + k = n\).

Theorem 5.41: Measure of a Dilation

Suppose \(t > 0\). If \(E \subseteq B_n\), then \(tE \subseteq B_n\) and \(\lambda_n(tE) = t^n\lambda_n(E)\)

where \(tE = \{tx: \forall x \in E\}\)

Definition 5.43: Open Unit Ball in \(\mathbb{R}^n\); \(B_n\)

The open unit ball in \(\mathbb{R}^n\) is denoted by \(\mathbf{B}_n\) and is defined by:

\[\mathbf{B}_n = \{(x_1, ..., x_n) \in \mathbb{R}^n : x^2_1 + .... + x_n^2 < 1\}\]

Definition 5.46: Partial Derivatives; \(D_1 f\) and \(D_2 f\)

Suppose \(G\) is an open subset of \(\mathbb{R}^2\) and \(f: G \rightarrow \mathbb{R}\) is a function. For \((x, y) \in G\), the partial derivatives \((D_1f)(x, y)\) and \(D_2f(x, y)\) are defined by:

\[(D_1f)(x, y) = \lim_{t\rightarrow 0} \frac{f(x+t, y) - f(x, y)}{t}\]

\[(D_2f)(x, y) = \lim_{t \rightarrow 0} \frac{f(x, y+t) - f(x, y)}{t}\]

If these limit exists.

Theorem 5.48: Equality of Mixed Partial Derivatives

Suppose \(G\) is an open subset of \(\mathbb{R}^2\) and \(f: G \rightarrow \mathbb{R}\) is a funciton s.t \(D_1 f\), \(D_2 f\), \(D_1 (D_2 f)\), \(D_2 (D_1 f)\) all exists and are continuous functions on \(G\). Then:

\[D_1(D_2 f) = D_2(D_1 f)\]