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. $(A\cup B)\times (C\cup D)\ne (A\times C)\cup (B\times D)$

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, $f:X\to [0,\mathrm{\infty }]$ 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:

$$L(f,P)=\sum _{j=1}^m\mu (A_j)\underset{A_j}{inf}f$$

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 $f:X\to [0,\mathrm{\infty }]$ is an $S$-measurable function. The integral of $f$ w.r.t $\mu$, denoted by $\int fd\mu$ is defined by:

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

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:

$$L({\chi }_E,P)=\mu (E)\ast 1+0\ast \mu (X/E)=\mu (E)$$

Thus,

$$\int {\chi }_{E}d\mu \ge \mu (E)$$

Let $P$ be an $S$-partition $A_1,...,A_m$ of $X$, then

$$\mu (A_j)\underset{A_j}{inf}{\chi }_E=\{\begin{array}{l}\mu (A_j),\text{if }{A}_j\subseteq E\\ 0,\text{ o. w }\end{array}$$

Then:

$$\sum _{j=1}^m\mu (A_j)\underset{A_j}{inf}{\chi }_E=\sum _{\{j:A_j\subseteq E\}}\mu (A_j)=\mu (\bigcup _{\{j:A_j\subseteq E\}}{A}_j)\le \mu (E)$$

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 $c_1,...,c_n\in [0,\mathrm{\infty }]$. Then:

$$\int (\sum _{k=1}^n{c}_k{\chi }_{E_k})d\mu =\sum _{k=1}^n{c}_k\mu (E_k)$$

Theorem 3.8: Integration is Order Preserving

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

$$\int fd\mu \le \int gd\mu$$

Proof of Theorem 3.8:

Suppose $P$ is a $S$-partition $A_1,...,A_n$ of $X$. Then:

$$\underset{A_i}{inf}f\le \underset{A_i}{inf}g⟹L(f,P)\le L(g,P)$$

Then for any $P$, we have:

$$\underset{P}{sup}L(f,P)\le \underset{P}{sup}L(g,P)$$

Theorem 3.9: Integral Via Simple Functions

Suppose $(X,S,\mu )$ is a measure space and $f:X\to [0,\mathrm{\infty }]$ is $S$-measurable. Then:

$$\int fd\mu =sup\{\sum _{j=1}^m{c}_j\mu (A_j):A_1,...,A_m\text{ are disjoint sets in }S\text{, }{c}_1,....,c_m\in [0,\mathrm{\infty })\text{, }f(x)\ge \sum _{j=1}^m{c}_j{\chi }_{A_j}(x)\text{ for every }x\in X\}$$

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 $0\le f_1\le f_2\le ...$ is an increasing sequence of $S$-measurable functions. Define $f:X\to [0,\mathrm{\infty }]$ by:

$$f(x)=\underset{k\to \mathrm{\infty }}{lim}{f}_k(x)$$

Then:

$$\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu =\int fd\mu$$

If:

$$f(x)=\underset{k\to \mathrm{\infty }}{lim}\sum _{n=1}^k{f}_n(x)$$

Then:

$$\sum _{n=1}^{\mathrm{\infty }}\int f_{n}d\mu =\int fd\mu$$

Proof of Theorem 3.11:

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

1. $\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu \le \int fd\mu$
2. $\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu \ge \int fd\mu$

We first prove 1:

Suppose $(X,S,\mu )$ is a measure space and $0\le f_1\le f_2\le ...$ is an increasing sequence of $S$-measurable functions. Define $f:X\to [0,\mathrm{\infty }]$ by:

$$f(x)=\underset{k\to \mathrm{\infty }}{lim}{f}_k(x)$$

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

$$\int f_{k}d\mu \le \int fd\mu ⟹\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu \le \int fd\mu$$

To prove 2, we need to use theorem 3.9:

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

$$f(x)\ge \sum _{j=1}^m{c}_j{\chi }_{A_j}(x)$$

Then:

$$\int fd\mu =\underset{\text{all disjoint sets in }S}{sup}\{\sum _{j=1}^m{c}_j\mu (A_j)\}$$

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

$$E_k=\{x\in X:f_k(x)\ge t\sum _{j=1}^m{c}_j{\chi }_{A_j}(x)\}$$

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

$$\mu (\underset{k\to \mathrm{\infty }}{lim}(E_k\cap A_j))=\mu (X\cap A_j)=\mu (A_j)$$

Then for all $x\in X$:

$$f_k(x)\ge t\sum _{j=1}^m{c}_j{\chi }_{A_j\cap E_k}(x)⟹\int f_{k}d\mu \ge \int t\sum _{j=1}^m{c}_j{\chi }_{A_j\cap E_k}d\mu ⟹\int f_{k}d\mu \ge t\sum _{j=1}^m{c}_j\mu (A_j\cap E_k)$$

Taking the limit to infinity:

$$\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu \ge \underset{k\to \mathrm{\infty }}{lim}t\sum _{j=1}^m{c}_j\mu (A_j\cap E_k)=t\sum _{j=1}^m{c}_j\mu (A_j)$$

Taking the limit $t\to 1$:

$$\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu \ge \sum _{j=1}^m{c}_j\mu (A_j)$$

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

$$\underset{k\to \mathrm{\infty }}{lim}{f}_k(x)\ge \underset{\text{all disjoint sets in }S}{sup}\{\sum _{j=1}^m{c}_j\mu (A_j)\}=\int fd\mu$$

Theorem 3.13: Integral-type Sums for Simple Functions

Suppose $(X,S,\mu )$ is a measurable space. Suppose $a_1,...,a_m,b_1,...,b_n\in [0,\mathrm{\infty }]$ and $A_1,...,A_m,B_1,...,B_n\in S$ s.t $\sum _{j=1}^m{a}_j{\chi }_{A_j}=\sum _{k=1}^n{b}_k{\chi }_{B_k}$. Then:

$$\sum _{j=1}^m{a}_j\mu (A_j)=\sum _{k=1}^n{b}_k\mu (B_k)$$

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 $c_1,...,c_n\in [0,\mathrm{\infty })$. Then:

$$\int (\sum _{k=1}^n{c}_k{\chi }_{E_k})d\mu =\sum _{k=1}^n{c}_k\mu (E_k)$$

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 $f,g:X\to [0,\mathrm{\infty }]$ are $S$-measurable functions. Then:

$$\int (f+g)d\mu =\int fd\mu +\int gd\mu$$

Proof of Theorem 3.16:

Suppose $f,g$ are simple functions, then:

$$\int (f+g)d\mu =\int (\sum _j{c}_j{\chi }_{E_j}+\sum _k{c}_k{\chi }_{E_k})d\mu =\sum _j{c}_j\mu (E_j)+\sum _k{c}_k\mu (E_k)=\int fd\mu +\int gd\mu$$

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:

$$\underset{k\to \mathrm{\infty }}{lim}{f}_k=f$$

$$\underset{k\to \mathrm{\infty }}{lim}{g}_k=g$$

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:

$$\int (f+g)d\mu =\underset{k\to \mathrm{\infty }}{lim}\int (f_k+g_k)d\mu =\underset{k\to \mathrm{\infty }}{lim}\int f_k\mu +\underset{k\to \mathrm{\infty }}{lim}\int g_{k}d\mu =\int fd\mu +\int gd\mu$$

Definition 3.17: $f^{+},f^{-}$

Suppose $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ is a function. Define function $f^{+}$ and $f^{-}$ from $X$ to $[0,\mathrm{\infty }]$ by:

$$f^{+}(x)=\{\begin{array}{l}{f}^{+}(x),\text{if }f(x)\ge 0\\ 0,\text{if }f(x)<0\end{array}$$

and

$$f^{-}(x)=\{\begin{array}{l}0,\text{if }f(x)\ge 0\\ -f^{-}(x),\text{if }f(x)<0\end{array}$$

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: $\int fd\mu$

Suppose $(X,S,\mu )$ is a measure space and $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ 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 $\int fd\mu$ is defined by:

$$\int fd\mu =\int f^{+}d\mu -\int f^{-}d\mu$$

The condition $\int |f|d\mu <\mathrm{\infty }$ is equivalent to the condition $\int f^{+}d\mu <\mathrm{\infty }$ and $\int f^{-}d\mu <\mathrm{\infty }$

Example 3.19

Suppose $\lambda$ is a Lebesgue measure on $\mathbb{R}$ and $f:\mathbb{R}\to \mathbb{R}$ is a function defined by:

$$f(x)=\{\begin{array}{l}1,\text{if }x\ge 0\\ 0,\text{if }x<0\end{array}$$ Then $\int f^{+}d\lambda =\mathrm{\infty }$ and $\int f^{-}d\lambda =\mathrm{\infty }$, so $\int fd\lambda$ is not defined.

Theorem 3.20: Integration is Homogeneous

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

$$\int cfd\mu =c\int fd\mu$$

Theorem 3.21: Additivity of Integration

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

$$\int (f+g)d\mu =\int fd\mu +\int gd\mu$$

Theorem 3.22: Integration is Order Preserving

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

Theorem 3.23: Absolute Value of Integral $\le$ Integral of Absolute Value

Suppose $(X,S,\mu )$ is a measure space and $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ is a function s.t $\int fd\mu$ is defined. Then:

$$|\int fd\mu |\le \int |f|d\mu$$

Limits of Integrals and Integrals of Limits

Definition 3.24: Integration on a Subset; ${\int }_{E}fd\mu$

Suppose $(X,S,\mu )$ is a measure space and $E\in S$. If $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ is an $S$-measurable function, then ${\int }_{E}fd\mu$ is defined by:

$${\int }_{E}fd\mu =\int {\chi }_{E}fd\mu$$

If the right side of the equation above is defined, o.w ${\int }_{E}fd\mu$ is undefined.

We can also think of ${\int }_{E}fd\mu$ 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, ${\int }_{X}fd\mu$ means the same as $\int fd\mu$.

Theorem 3.25: Bounded an Integral

Suppose $(X,S,\mu )$ is a measure space, $E\in S$ and $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ is a function s.t ${\int }_{E}fd\mu$ is defined. Then:

$$|{\int }_{E}fd\mu |\le \mu (E)\underset{E}{sup}|f|$$

Proof Theorem 3.25:

Let $c=\underset{E}{sup}|f|$:

$$|{\int }_{E}fd\mu |=|\int {\chi }_{E}fd\mu |\le \int {\chi }_E|f|d\mu \le \int {\chi }_{E}cd\mu =c\mu (E)$$

Theorem 3.26: Bounded Convergence Theorem

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

$$|f_k(x)\le c|$$

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

$$\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu =\int fd\mu$$

Proof of Theorem 3.26:

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

$$|\int f_{k}d\mu -\int fd\mu |=|{\int }_{X/E}(f_k-f)d\mu +{\int }_E(f_k-f)d\mu |=|{\int }_{X/E}{f}_{k}d\mu -{\int }_{X/E}fd\mu +{\int }_E(f_k-f)d\mu |$$

Then by theorem 3.23:

$$|{\int }_{X/E}{f}_{k}d\mu -{\int }_{X/E}fd\mu +{\int }_E(f_k-f)d\mu |\le {\int }_{X/E}|f_k|d\mu +{\int }_{X/E}|f|d\mu +{\int }_E|f_k-f|d\mu |$$

Since $|f_k(x)|\le c\mathrm{\forall }x\in X$ and by theorem 3.25:

$${\int }_{X/E}|f_k|d\mu +{\int }_{X/E}|f|d\mu +{\int }_E|f_k-f|d\mu |\le \int c{\chi }_{X/E}d\mu +\int c{\chi }_{X/E}d\mu +{\int }_E|f_k-f|d\mu |\le \frac{ϵ}{2}+\mu (E)\underset{E}{sup}|f_k-f|$$

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

$$\underset{k\to \mathrm{\infty }}{lim}|\int f_{k}d\mu -\int fd\mu |=0$$

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, $g:X\to [0,\mathrm{\infty }]$ is $S$-measurable, and $\int gd\mu <\mathrm{\infty }$. Then for every $ϵ>0$, there exists $\delta >0$ s.t:

$${\int }_{B}gd\mu <ϵ$$

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, $g:X\to [0,\mathrm{\infty }]$ is $S$-measurable, and $\int gd\mu <\mathrm{\infty }$. Then for every $ϵ>0$, there exists $E\in S$ s.t $\mu (E)<\mathrm{\infty }$ and

$${\int }_{X/E}gd\mu <ϵ$$

Theorem 3.31: Dominated Convergence Theorem

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

$$\underset{k\to \mathrm{\infty }}{lim}{f}_k(x)=f(x)$$

for almost every $x\in X$. If there exists an $S$-measurable function $g:X\to [0,\mathrm{\infty }]$ s.t:

$$\int gd\mu <\mathrm{\infty }$$

and

$$|f_k(x)|\le g(x)$$

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

$$\underset{k\to \mathrm{\infty }}{lim}\int f_{k}d\mu =\int fd\mu$$

Theorem 3.34: Riemann Integrable $⟺$ Continuous Almost Everywhere

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

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

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:

$${\int }_a^{b}f={\int }_{[a,b]}fd\lambda$$

Definition 3.39: ${\int }_a^{b}f$

We previously defined the notation ${\int }_a^{b}f$ to mean the Riemann integral of $f$, now we redefine ${\int }_a^{b}f$ to denote the Lebesgue integral.

Suppose $-\mathrm{\infty }\le a<b\le \mathrm{\infty }$ and $f:(a,b)\to \mathbb{R}$ is Lebesgue measurable. Then:

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

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

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

$$\Vert f{\Vert }_1=\int |f|d\mu$$

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

$$L^1(\mu )=\{f:ff\text{ is an }S\text{-measurable function from }X\to \mathbb{R}\text{ and }\Vert f{\Vert }_1<\mathrm{\infty }\}$$

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 $\mu (E_k)<\mathrm{\infty }$.

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:

• $\Vert f{\Vert }_1\ge 0$
• $\Vert f{\Vert }_1=0$ IFF $f(x)=0$ for almost every $x\in X$
• $\Vert cf{\Vert }_1=|c|\Vert f{\Vert }_1,\mathrm{\forall }c\in \mathbb{R}$
• $\Vert f+g{\Vert }_1\le \Vert f{\Vert }_1+\Vert g{\Vert }_1$

Theorem 3.44: Approximation by Simple Functions

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

$$\Vert f-g{\Vert }_1<ϵ$$

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

• 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 $\Vert f{\Vert }_1$ 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 $g:\mathbb{R}\to \mathbb{R}$ 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 $ϵ>0$, there exists a step function $g\in L^1(\mathbb{R})$ s.t:

$$||f-g{\Vert }_1<ϵ$$

Theorem 3.48: Approximation by Continuous Functions

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

$$\Vert f-g{\Vert }_1<ϵ$$

and $\{x\in \mathbb{R}:g(x)\ne 0\}$ 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:

$$\mu (\{x\in X:|h(x)|\ge c\})\le \frac{1}{c}\Vert h{\Vert }_1$$

for every $c>0$.

Proof of Theorem 4.1:

Suppose $c>0$. Then:

$$\begin{array}{rl}\mu (\{x\in X:|h(x)|\ge c\})& ={\int }_{\{x\in X:|h(x)|\ge c\}}1d\mu \\ & =\frac{1}{c}{\int }_{\{x\in X:|h(x)|\ge c\}}cd\mu \\ & \le \frac{1}{c}{\int }_{\{x\in X:|h(x)|\ge c\}}|h|d\mu \\ & \le \frac{1}{c}{\int }_{\{x\in X:|h(x)|\ge c\}}|h|d\mu +\frac{1}{c}{\int }_{\{x\in X:|h(x)|<c\}}|h|d\mu \\ & =\frac{1}{c}\Vert h{\Vert }_1\end{array}$$

Definition 4.2: $3$ times a Bounded Nonempty Open Interval

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

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

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:

$$I_1\cup ...\cup I_n\subseteq (3\ast I_{k_1})\cup ....\cup (3\ast I_{k_n})$$

Example 4.5:

Suppose $n=4$ and

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

Then:

$$3\ast I_1=(-10,20),3\ast I_2=(3,21),3\ast I_3=(6,30),3\ast I_4=(11,41)$$

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

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; $h^{\ast }$

Suppose $h:\mathbb{R}\to \mathbb{R}$ is a Lebesgue measurable function. Then the Hardy-Littlewood maximal function of $h$ is the function $h^{\ast }:\mathbb{R}\to [0,\mathrm{\infty }]$ defined by:

$$h^{\ast }(b)=\underset{t>0}{sup}\frac{1}{2t}{\int }_{b-t}^{b+t}|h|=\underset{t>0}{sup}\frac{1}{2t}{\int }_{(b+t,b-t)}|h|d\lambda$$

Where $h^{\ast }$ is a Borel measurable function.

In other words, $h^{\ast }(b)$ 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:

$$|\{b\in \mathbb{R}:h^{\ast }(b)>c\}|\le \frac{3}{c}\Vert h{\Vert }_1$$

Derivatives of Integrals

Theorem 4.10: Lebesgue Differentiation Theorem, First Version

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

$$\underset{t\to 0}{lim}\frac{1}{2t}{\int }_{b-t}^{b+t}|f-f(b)|=0$$

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 $g:I\to \mathbb{R}$ 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:

$$g^{\prime }(b)=\underset{t\to 0}{lim}\frac{g(b+t)-g(b)}{t}$$

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 $g:\mathbb{R}\to \mathbb{R}$ by:

$$g(x)={\int }_{-\mathrm{\infty }}^{x}f$$

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 $t\ne 0$, then:

$$\begin{array}{rl}|\frac{g(b+t)-g(b)}{t}-f(b)|& =|\frac{{\int }_{-\mathrm{\infty }}^{b+t}f-{\int }_{-\mathrm{\infty }}^{b}f}{t}-f(b)|\\ & =|\frac{{\int }_b^{b+t}f}{t}-f(b)|\\ & =|\frac{{\int }_b^{b+t}f}{t}-\frac{{\int }_b^{b+t}f(b)}{t}|\\ & =|\frac{{\int }_b^{b+t}f-f(b)}{t}|\\ & \le \underset{\{x\in \mathbb{R}:|x-b|<|t|\}}{sup}|f(x)-f(b)|\end{array}$$

Since $f$ is continuous at $b$, for all $ϵ>0$, there exists $\delta >0$, s.t $|x-b|<\delta ⟹|f(x)-f(b)|<ϵ$, if we let $|t|\to 0$, s.t $|t|<\delta ,\mathrm{\forall }\delta >0$, then:

$$\underset{t\to 0}{lim}\underset{\{x\in \mathbb{R}:|x-b|<|t|\}}{sup}|f(x)-f(b)|<ϵ$$

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

Theorem 4.19: Lebesgue Differentiation Theorem, Second Version

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

$$g(x)={\int }_{-\mathrm{\infty }}^{x}f$$

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:

$$f(b)=\underset{t\to 0}{lim}\frac{1}{2t}{\int }_{b-t}^{b+t}f$$

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:

$$\underset{t\to 0}{lim}\frac{|E\cap (b-t,b+t)|}{2t}$$

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

Example 4.23

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

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:

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

Example 5.5:

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

$$[A\times B{]}_a=\{\begin{array}{ll}B,& \text{if }a\in A\\ \mathrm{\varnothing },& \text{if }a\notin A\end{array}$$

and

$$[A\times B{]}^b=\{\begin{array}{ll}A,& \text{if }b\in B\\ \mathrm{\varnothing },& \text{if }b\notin B\end{array}$$

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

$$E_a\in T,\mathrm{\forall }a\in X,E^b\in S,\mathrm{\forall }b\in Y$$

In other words, Let $\epsilon$ denotes the collection of subsets $E$ of $X\times Y$ for which the conclusion of this result holds. Then $\epsilon$ 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 $f:X\times Y\to \mathbb{R}$ is a function. Then for $a\in X$ and $b\in Y$, the cross section functions $f_a:Y\to \mathbb{R}$ and $f^b:X\to \mathbb{R}$ are defined by:

$$f_a(y)=f(a,y),\mathrm{\forall }y\in Y$$

and

$$f^b(x)=f(x,b),\mathrm{\forall }x\in X$$

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 $f:X\times Y\to \mathbb{R}$ is an $S\otimes T$-measurable function. Then:

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

and

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

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:

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

Then,

$$\begin{array}{rl}y\in (f_a{)}^{-1}(D)& ⟺f(a,y)\in D\\ & ⟺(a,y)\in f^{-1}(D)\\ & ⟺y\in [f^{-1}(D){]}_a\end{array}$$

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

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

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

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. $\mathrm{\varnothing }\in A$.
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: $$\bigcup _{k=1}^{\mathrm{\infty }}{E}_k\in M$$
• If $E_1\supseteq E_2\supseteq ...$ is an decreasing sequence of sets in $M$, then: $$\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\in M$$

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 $\mu (X)<\mathrm{\infty }$.
• 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: $$X=\bigcup _{k=1}^{\mathrm{\infty }}{X}_k,\mu (X_k)<\mathrm{\infty },\mathrm{\forall }k\in {\mathbb{Z}}^{+}$$

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. $x⟼v(E_x)$ is an $S$-measurable function on $X$ (Think of the function as $f(x)=v(E_x)$).
2. $y⟼\mu (E^y)$ 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 $g:X\to [-\mathrm{\infty },\mathrm{\infty }]$ is a function. The notation:

$$\int g(x)d\mu (x):=\int gd\mu$$

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 $f:X\times Y\to \mathbb{R}$ 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:

$$x⟼{\int }_{Y}f(x,y)dv(y)$$

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 $f:X\times Y\to [0,\mathrm{\infty }]$ is $S\otimes T$ measurable. Then:

$$x⟼{\int }_{Y}f(x,y)dv(y)\text{ is an }S\text{-measurable function on }X$$

$$y⟼{\int }_{X}f(x,y)d\mu (x)\text{ is an }T\text{-measurable function on }Y$$

and

$${\int }_{X\times Y}fd(\mu \times v)={\int }_X{\int }_{Y}f(x,y)dv(y)d\mu (x)={\int }_Y{\int }_{X}f(x,y)d\mu (x)dv(y)$$

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:

$$\sum _{j=1}^{\mathrm{\infty }}\sum _{k=1}^{\mathrm{\infty }}{x}_{j,k}=\sum _{k=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{x}_{j,k}$$

Theorem 5.32: Fubini's Theorem

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

$${\int }_Y|f(x,y)|dv(y)<\mathrm{\infty }\text{ for almost every }x\in X$$

and

$${\int }_Y|f(x,y)|d\mu (x)<\mathrm{\infty }\text{ for almost every }y\in Y$$

Furthermore:

$$x⟼{\int }_{Y}f(x,y)dv(y)\text{ is an }S\text{-measurable function on }X$$

$$y⟼{\int }_{X}f(x,y)d\mu (x)\text{ is an }T\text{-measurable function on }Y$$

And:

$${\int }_{X\times Y}fd(\mu \times v)={\int }_X{\int }_{Y}f(x,y)dv(y)d\mu (x)={\int }_Y{\int }_{X}f(x,y)d\mu (x)dv(y)$$

Definition 5.34: Region Under the Graph; $U_f$

Suppose $X$ is a set and $f:X\to [0,\mathrm{\infty }]$ is a function. Then the region under the graph of $f$, denoted $U_f$ is defined by:

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

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

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

$$(\mu \times \lambda )(U_f)={\int }_{X}fd\mu ={\int }_{(0,\mathrm{\infty })}\mu (\{x\in X:t<f(x)\})d\lambda (t)$$

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: $$B(x,\delta )=\{y\in {\mathbb{R}}^n:\Vert y-x{\Vert }_{\mathrm{\infty }}<\delta \}$$
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 $sup\{\Vert a{\Vert }_{\mathrm{\infty }}:a\in A\}<\mathrm{\infty }$.
7. ${\mathbb{R}}^m\times {\mathbb{R}}^n$ is identified with ${\mathbb{R}}^{m+n}$, note that ${\mathbb{R}}^2\times {\mathbb{R}}^1\ne {\mathbb{R}}^3$ 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 $tE=\{tx:\mathrm{\forall }x\in E\}$

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:

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

Definition 5.46: Partial Derivatives; $D_{1}f$ and $D_{2}f$

Suppose $G$ is an open subset of ${\mathbb{R}}^2$ and $f:G\to \mathbb{R}$ is a function. For $(x,y)\in G$, the partial derivatives $(D_{1}f)(x,y)$ and $D_{2}f(x,y)$ are defined by:

$$(D_{1}f)(x,y)=\underset{t\to 0}{lim}\frac{f(x+t,y)-f(x,y)}{t}$$

$$(D_{2}f)(x,y)=\underset{t\to 0}{lim}\frac{f(x,y+t)-f(x,y)}{t}$$

If these limit exists.

Theorem 5.48: Equality of Mixed Partial Derivatives

Suppose $G$ is an open subset of ${\mathbb{R}}^2$ and $f:G\to \mathbb{R}$ 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:

$$D_1(D_{2}f)=D_2(D_{1}f)$$