measure – Mathtuition88

Interesting Measure and Integration Question

Let $(\Omega,\mathcal{A},\mu)$ be a measure space. Let $f\in L^p$ and $\epsilon>0$ . Prove that there exists a set $E\in\mathcal{A}$ with $\mu(E)<\infty$ , such that $\int_{E^c} |f|^p<\epsilon$ .

Solution:

The solution strategy is to use simple functions (common tactic for measure theory questions).

Let $0\leq\phi\leq |f|^p$ be a simple function such that $\int_\Omega (|f|^p-\phi)\ d\mu<\epsilon$ .

Consider the set $E=\{\phi>0\}$ . Note that $\int_\Omega \phi\ d\mu\leq\int_\Omega |f|^p\ d\mu<\infty$ . Hence each nonzero value of $\phi$ can only be on a set of finite measure. Since $\phi$ has only finitely many values, $\mu(E)<\infty$ .

Then,

$\begin{aligned} \int_{E^c}|f|^p\ d\mu&=\int_{E^c} (|f|^p-\phi)\ d\mu +\int_{E^c}\phi\ d\mu\\ &\leq \int_\Omega (|f|^p-\phi)\ d\mu+0\\ &<\epsilon \end{aligned}$

Measure that is absolutely continuous with respect to mu

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Let $(X,\mathcal{M},\mu)$ be a measure space, and let $f:X\to [0,\infty]$ be a measurable function. Define the map $\lambda:\mathcal{M}\to[0,\infty]$ , $\lambda(E):=\int_X \chi_E f d\mu$ , where $\chi_E$ denotes the characteristic function of $E$ .

(a) Show that $\lambda$ is a measure and that it is absolutely continuous with respect to $\mu$ .

(b) Show that for any measurable function $g:X\to[0,\infty]$ , one has $\int_X g d\lambda=\int_X gf d\mu$ in $[0,\infty]$ .

Proof: For part (a), we routinely check that $\lambda$ is indeed a measure.

$\lambda(\emptyset)=\int_X \chi_\emptyset f d\mu=\int_X 0 d\mu=0$ . Let $E_i$ be mutually disjoiint measurable sets.

$\begin{aligned} \lambda(\cup_{i=1}^\infty E_i)&=\int_X \chi_{\cup_{i=1}^\infty E_i} f d\mu\\ &=\int_X (\sum_{i=1}^\infty \chi_{E_i}) f d\mu\\ &=\sum_{i=1}^\infty \int_X \chi_{E_i} f d\mu\\ &=\sum_{i=1}^\infty \lambda (E_i) \end{aligned}$

If $\mu (E)=0$ , then $\chi_{E} f=0$ a.e., thus $\lambda (E)=0$ . Therefore $\lambda\ll\mu$ .

(b) We note that when $g$ is a characteristic function, i.e. $g=\chi_E$ ,

$\begin{aligned} \int_X g d\lambda&=\int_X \chi_E d\lambda\\ &=\lambda (E)\\ &=\int_X \chi_E f d\mu\\ &=\int_X gf d\mu \end{aligned}$

Hence the equation holds. By linearity, we can see that the equation holds for all simple functions. Let $(\psi_n)$ be a sequence of simple functions such that $\psi_n\uparrow g$ . Then by the Monotone Convergence Theorem, $\lim_{n\to\infty} \int \psi_n d\lambda=\int g d\lambda$ .

Note that $\psi_n f\uparrow gf$ , thus by MCT, $\lim_{n\to\infty}\int\psi_n f d\mu=\int g f d\mu$ . Note that $\int \psi_n d\lambda=\int \psi_n f d\mu$ . Hence, $\int g d\lambda=\int gf d\mu$ , and we are done.

Markov Inequality + PSLE One Dollar Question

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Markov inequality is a useful inequality that gives a rough upper bound of the measure of a set in terms of an integral. The precise statement is: Let $f$ be a nonnegative measurable function on $\Omega$ . The Markov inequality states that for all $K>0$ , $\displaystyle \mu\{x\in\Omega:f(x)\geq K\}\leq\frac{1}{K}\int fd\mu$ .

The proof is rather neat and short. Let $E_K:=\{x\in\Omega: f(x)\geq K\}$ Then,

$\begin{aligned} \int f d\mu &\geq \int_{E_K} fd\mu\\ &\geq \int_{E_K}K d\mu\\ &=K \mu(E_K) \end{aligned}$

Therefore, $\mu(E_K)\leq\frac{1}{K}\int fd\mu$ .

Necessary and Sufficient Condition for Integrability in finite measure space

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Let $(\Omega, \Sigma, \mu)$ be a finite measure space. Suppose that $f\geq 0$ is a measurable function on $\Omega$ . Let $E_n:=\{\omega\in\Omega:f(\omega)\geq n\}$ for each $n\in \mathbb{N}\cup\{0\}$ . Show that $f$ is integrable if and only if $\sum_{n=0}^\infty \mu(E_n)<\infty$ .

This proof has a cute solution that is potentially very short. We will elaborate more on this proof. Other approaches include using Markov’s Inequality / Chebyshev’s inequality.

Proof: Consider $g=\sum_{n=1}^\infty I_{E_n}$ .

Note that for each $n\geq 0, n\in\mathbb{N}$ on $E_n\setminus E_{n+1}$ , $g=n$ , while $n\leq f<n+1$ . Therefore $g\leq f<g+1$ on $\Omega$ .

Integrating with respect to $\mu$ , we get $\int_{\Omega} gd\mu\leq\int_{\Omega} fd\mu<\int_{\Omega} g+1d\mu$ .

$\displaystyle\boxed{\sum_{n=1}^\infty \mu (E_n)\leq \int_\Omega f d\mu<\sum_{n=1}^\infty\mu(E_n)+\mu(\Omega)}$

(=>) Now assuming f is integrable, i.e. $\int fd\mu<\infty$ , we have $\sum_{n=1}^\infty \mu(E_n)<\infty$ . $\mu(E_0)=\mu(\Omega)<\infty$ . Therefore $\sum_{n=0}^\infty\mu(E_n)<\infty$ .

(<=) Conversely, if $\sum_{n=0}^\infty\mu(E_n)=\sum_{n=1}^\infty \mu(E_n)+\mu(\Omega)<\infty$ , then $\int_\Omega f d\mu<\infty$ .

We are done.

Note: For a more rigorous proof of $\int gd\mu=\sum_{n=1}^\infty \mu (E_n)$ we can use MCT (Monotone Convergence Theorem).

Let $g_k=\sum_{n=1}^k I_{E_n}$ . Then $g_k\uparrow g$ . By MCT, $\int gd\mu=\lim_{k\to\infty} \int g_k d\mu=\lim_{k\to\infty} \sum_{n=1}^k \mu (E_n)=\sum_{n=1}^\infty \mu(E_n)$ .

Interesting Analysis Question (Measure Theory)

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Let $f\geq 0$ be a measurable function with $\int f d\mu<\infty$ . Show that for any $\epsilon>0$ , there exists a $\delta(\epsilon)>0$ such that for any measurable set $E\in\mathcal{A}$ with $\mu(E)<\delta(\epsilon)$ , we have $\int_E f d\mu<\epsilon$ .

Proof: For $M>0$ , we define $f_M(x)=\min (f(x),M)\leq M$ , for all $x\in \Omega$ .

Then $f=f_M+(f-f_M)$ .

Let $\delta=\epsilon/2M$ . Then for any $E\in\mathcal{A}$ with $\mu(E)<\delta$ ,

$\begin{aligned}\int_E f_M d\mu &\leq \int_E M d\mu\\ &=\mu (E)M\\ &<\delta M\\ &=(\epsilon/2M)M\\ &=\epsilon/2 \end{aligned}$

Note that $f_M\uparrow f$ . By Monotone Convergence Theorem,

$\int f d\mu=\lim_{M\to\infty}\int f_M d\mu$ .

Therefore $\lim_{M\to\infty}\int f-f_M d\mu=0$ .

We can choose $M$ sufficiently large such that

$\int_E f-f_M d\mu \leq \int f-f_M d\mu <\epsilon/2$ .

Then

$\begin{aligned} \int_E f d\mu&=\int_E f_M d\mu+\int_E f-f_M d\mu\\ &<\epsilon/2+\epsilon/2\\ &=\epsilon \end{aligned}$

We are done!

Measure Theory: What does a.e. (almost everywhere) mean

Source: Elements of Integration by Professor Bartle

Students studying Mathematical Analysis, Advanced Calculus, or probability would sooner or later come across the term a.e. or “almost everywhere”.

In layman’s terms, it means that the proposition (in the given context) holds for all cases except for a certain subset which is very small. For instance, if f(x)=0 for all x, and g(x)=0 for all nonzero x, but g(0)=1, the function f and g would be equal almost everywhere.

For formally, a certain proposition holds $\mu$ -almost everywhere if there exists a subset $N\in \mathbf{X}$ with $\mu (N)=0$ such that the proposition holds on the complement of N. $\mu$ is a measure defined on the measure space $\mathbf{X}$ , which is discussed in a previous blog post: What is a Measure.

Two functions $f, g$ are said to be equal $\mu$ -almost everywhere when $f(x)=g(x)$ when $x\notin N$ , for some $N\in X$ with $\mu (N)=0$ . In this case we would often write $f=g$ , $\mu$ -a.e.

Similarly, this notation can be used in the case of convergence, for example $f=\lim f_n$ , $\mu$ -a.e.

The idea of “almost everywhere” is useful in the theory of integration, as there is a famous Theorem called “Lebesgue criterion for Riemann integrability”.

(From Wikipedia)

A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). This is known as the Lebesgue’s integrability condition or Lebesgue’s criterion for Riemann integrability or the Riemann—Lebesgue theorem.^[4] The criterion has nothing to do with the Lebesgue integral. It is due to Lebesgue and uses his measure zero, but makes use of neither Lebesgue’s general measure or integral.

Reference book:

Lemma on Measure of Increasing Sequences in X

(Continued from https://mathtuition88.com/2015/06/20/what-is-a-measure-measure-theory/)

Lemma: Let $\mu$ be a measure defined on a $\sigma$ -algebra X.

(a) If ( $E_n$ ) is an increasing sequence in X, then

$\mu (\bigcup_{n=1}^\infty E_n )=\lim \mu (E_n)$

(b) If $F_n$ ) is a decreasing sequence in X and if $\mu (F_1)<\infty$ , then

$\mu (\bigcap_{n=1}^\infty F_n )=\lim \mu (F_n )$

Note: An increasing sequence of sets ( $E_n$ ) means that for all natural numbers n, $E_n \subseteq E_{n+1}$ . A decreasing sequence means the opposite, i.e. $E_n \supseteq E_{n+1}$ .

Proof: (Elaboration of the proof given in Bartle’s book)

(a) First we note that if $\mu (E_n) = \infty$ for some n, then both sides of the equation are $\infty$ , and inequality holds. Henceforth, we can just consider the case $\mu (E_n)<\infty$ for all n.

Let $A_1 = E_1$ and $A_n=E_n \setminus E_{n-1}$ for n>1. Then $(A_n)$ is a disjoint sequence of sets in X such that

$E_n=\bigcup_{j=1}^n A_j$ , $\bigcup_{n=1}^\infty E_n = \bigcup_{n=1}^\infty A_n$

Since $\mu$ is countably additive,

$\mu (\bigcap_{n=1}^\infty E_n) = \sum_{n=1}^\infty \mu (A_n)$ (since ( $A_n$ ) is a disjoint sequence of sets)

$=\lim_{m\to\infty} \sum_{n=1}^m \mu (A_n)$

By an earlier lemma $\mu (F\setminus E)=\mu (F)-\mu (E)$ , we have that $\mu (A_n)=\mu (E_n)-\mu (E_{n-1})$ for n>1, so the finite series on the right side telescopes to become

$\sum_{n=1}^m \mu (A_n)=\mu (E_m)$

Thus, we indeed have proved (a).

For part (b), let $E_n=F_1 \setminus F_n$ , so that $(E_n)$ is an increasing sequence of sets in X.

We can then apply the results of part (a).

$\begin{aligned} \mu (\bigcup_{n=1}^\infty E_n) &=\lim \mu (E_n)\\ &=\lim [\mu (F_1)-\mu (F_n)]\\ &=\mu (F_1) -\lim \mu (F_n) \end{aligned}$

Since we have $\bigcup_{n=1}^\infty E_n =F_1 \setminus \bigcap_{n=1}^{\infty} F_n$ , it follows that

$\mu (\bigcup_{n=1}^\infty E_n) =\mu (F_1)-\mu (\bigcap_{n=1}^\infty F_n)$

Comparing the above two equations, we get our desired result, i.e. $\mu (\bigcap_{n=1}^\infty F_n) = \lim \mu (F_n)$ .

Reference:

The Elements of Integration and Lebesgue Measure

What is a Measure? (Measure Theory)

In layman’s terms, “measures” are functions that are intended to represent ideas of length, area, mass, etc. The inputs for the measure functions would be sets, and the output would be a real value, possibly including infinity.

It would be desirable to attach the value 0 to the empty set $\emptyset$ and measures should be additive over disjoint sets in X.

Definition (from Bartle): A measure is an extended real-valued function $\mu$ defined on a $\sigma$ -algebra X of subsets of X such that
(i) $\mu (\emptyset)=0$
(ii) $\mu (E) \geq 0$ for all $E\in \mathbf{X}$
(iii) $\mu$ is countably additive in the sense that if $(E_n)$ is any disjoint sequence ( $E_n \cap E_m =\emptyset\ \text{if }n\neq m$ ) of sets in X, then

$\displaystyle \mu(\bigcup_{n=1}^\infty E_n )=\sum_{n=1}^\infty \mu (E_n)$ .

If a measure does not take on $+\infty$ , we say it is finite. More generally, if there exists a sequence $(E_n)$ of sets in X with $X=\cup E_n$ and such that $\mu (E_n) <+\infty$ for all n, then we say that $\mu$ is $\sigma$ -finite. We see that if a measure is finite implies it is $\sigma$ -finite, but not necessarily the other way around.

Examples of measures

(a) Let X be any nonempty set and let X be the $\sigma$ -algebra of all subsets of X. Let $\mu_1$ be definied on X by $\mu_1 (E)=0$ , for all $E\in\mathbf{X}$ . We can see that $\mu_1$ is finite and thus also $\sigma$ -finite.

Let $\mu_2$ be defined by $\mu_2 (\emptyset) =0$ , $\mu_2 (E)=+\infty$ if $E\neq \emptyset$ . $\mu_2$ is an example of a measure that is neither finite nor $\sigma$ -finite.

The most famous measure is definitely the Lebesgue measure. If X=R, and X=B, the Borel algebra, then (shown in Bartle’s Chapter 9) there exists a unique measure $\lambda$ defined on B which coincides with length on open intervals. I.e. if E is the nonempty interval (a,b), then $\lambda (E)=b-a$ . This measure is usually called Lebesgue measure (or sometimes Borel measure). It is not a finite measure since $\lambda (\mathbb{R})=\infty$ . But it is $\sigma$ -finite since any interval can be broken down into a sequence of sets ( $E_n$ ) such that $\mu (E_n)<\infty$ for all n.

Source: The Elements of Integration and Lebesgue Measure

Functions between Measurable Spaces

Sometimes, it is desirable to define measurability for a function f from one measurable space (X,X) into another measurable space (Y,Y). In this case one can define f to be measurable if and only if the set $f^{-1} (E)=\{x\in X: f(x) \in E\}$ belongs to X for every set E belonging to Y.

This definition of measurability appears to differ from Definition 2.3 (earlier in the book), but Definition 2.3 is in fact equivalent to this definition in the case that Y=R and Y=B.

First, lets recap what is Definition 2.3:

A function f on X to R is said to be X-measurable (or simply measurable) if for every real number $\alpha$ the set $\{x\in X:f(x)>\alpha\}$ belongs to X.

Let (X,X) be a measurable space and f be a real-valued function defined on X. Then f is X-measurable if and only if $f^{-1 }(E)\in X$ for every Borel set E.