Some Convergence Concepts

This result concerns uniqueness.

Suppose that $f_{n} \xrightarrow {\mu } f$ and $g_{n} \xrightarrow {\mu } g$.

Suppose that $f_{n} \xrightarrow {\text{ a.e. }} f$ and $g_{n} \xrightarrow {\text{ a.e. }} g$.

Suppose for each $n \in \mathbb {N}$ that $f_{n}=g_{n} \mu$-a. e..

Prove the following statements connecting convergence in $\mu$-measure with convergence $\mu$-a. e.

Suppose that $\varphi : \mathbb {R} \rightarrow \mathbb {R}$ is continuous.

Let $\Omega = [0, 1]$, let $\mathcal{F} = \left\{ B \subseteq [0,1] : B \in \mathcal{L}\right\}$, and let $\mu$ denote the restriction of Lebesgue measure $\overline{\lambda }$ to $\mathcal{F}$. Let $A_{1}=[0,1]$, $A_{2}=[0, \frac{1}{2}]$, $A_{3}=[\frac{1}{2}, 1]$, $A_{4}=[0, \frac{1}{4}]$, $A_{5}=[\frac{1}{4}, \frac{1}{2}]$, $A_{6}=[\frac{1}{2}, \frac{3}{4}]$, $A_{7}=[\frac{3}{4}, 1]$, $A_{8}=[0, \frac{1}{8}]$, etc. On $\Omega$, define $f_{n}=\; \mathbb {1}_{A_{n}}$ for each $n \in \mathbb {N}$.

Consider the measure space $(\mathbb {R}, \mathcal{L}, \overline{\lambda })$, and let $f_{n}(\omega )=\; \mathbb {1}_{[n, n+1]}(\omega )$ for each $n \in \mathbb {N}$ and $\omega \in \mathbb {R}$.

If $\mu (\left\{ \omega \in A : \left|f_{n}(\omega )-f(\omega )\right|>\epsilon \text{ for infinitely many } n\right\} )=0$ for each $\epsilon >0$, then $f_{n} \rightarrow f$ $\mu$-a.e.

We have assumed throughout Section 5.6 that the functions $f, f_{1}, f_{2}, \ldots$ have domain $A$ and are finite-valued. We may generalize the previous definitions and discussion to the case where each of $f, f_{1}, f_{2}, \ldots$ is defined $\mu$-a.e.on $A$ and each of $f, f_{1}, f_{2}, \ldots$ is finite $\mu$-a.e.on $A$. Let $B$ denote the set of $\omega \in A$ such that (i) $f(\omega )$ is defined and finite, and (ii) $f_{n}(\omega )$ is defined and finite for each $n \in \mathbb {N}$. Then

and $A - B$ is a countable union of $\mu$-null sets, so $\mu (A-B)=0$. We may therefore decompose $A$ into the two disjoint $\mathcal{F}$-sets $B$ and $A-B$, and think of $g, g_{1}, g_{2}, \ldots$ as denoting $\mathcal{F}/\mathfrak {B}^{*}$-measurable functions with domain $B$ and range contained in $\mathbb {R}$, where $g, g_{1}, g_{2}, \ldots$ denote the restrictions of $f, f_{1}, f_{2}, \ldots$ (respectively) from their respective domains to $B$. We say that $f_{n} \rightarrow f$ $\mu$-a.e.on $A$ iff $\mu (\left\{ \omega \in B : g_{n}(\omega ) \not\rightarrow g(\omega )\right\} )=0$, i.e., iff $g_{n} \rightarrow g$ $\mu$-a.e.on $B$; and we say that $f_{n} \rightarrow f$ in $\mu$-measure on $A$ iff $g_{n} \rightarrow g$ in $\mu$-measure on $B$, i.e., iff for every $\epsilon >0$ and $\delta >0$ we have $\lim_{n} \mu (\left\{ \omega \in B : \left|g_{n}(\omega )-g(\omega )\right|>\delta \right\} )=0$.

State and prove Claim 2 in this more general setting. Mimic the spirit of the example given earlier for Claim 1's own generalization: Claim 1 states that if $(\Omega , \mathcal{F}, \mu )$ denotes a measure space and $A \in \mathcal{F}-\left\{ \varnothing \right\}$, and if $f, f_{1}, f_{2}, \ldots$ are $\mathcal{F}/\mathfrak {B}^{*}$-measurable functions defined and finite $\mu$-a.e.on $A$ such that $f_{n} \rightarrow f$ in $\mu$-measure, then there exists a subsequence $\left\{ f_{n_{k}}\right\}_{k=1}^{\infty }$ of $\left\{ f_{n}\right\}_{n=1}^{\infty }$ such that $f_{n_{k}} \rightarrow f$ $\mu$-a.e.on $A$; this is proved by applying the original (everywhere-defined, finite-valued) version of Claim 1 to $g, g_{1}, g_{2}, \ldots$ on $B$ to get a subsequence $\left\{ g_{n_{k}}\right\}$ with $g_{n_{k}} \rightarrow g$ $\mu$-a.e.on $B$, which by definition means $f_{n_{k}} \rightarrow f$ $\mu$-a.e.on $A$.

Claim 2 (the result to be generalized here): Let $(\Omega , \mathcal{F}, \mu )$ denote a measure space with $\mu (\Omega )<+\infty$. Pick nonempty $A \in \mathcal{F}$. Suppose that (i) each of $f, f_{1}, f_{2}, \ldots$ is $\mathcal{F}/\mathfrak {B}^{*}$-measurable with domain $A$ and range contained in $\mathbb {R}$, and (ii) $f_{n} \rightarrow f$ $\mu$-a.e.on $A$. Then, given any $\epsilon >0$ and $\delta >0$, we may find a set $B \in \mathcal{F}$ and $n_{0} \in \mathbb {N}$ such that $\mu (B^{c})<\epsilon$ and $\left|f_{n}(\omega )-f(\omega )\right|<\delta$ for every $\omega \in B$ and $n \geqslant n_{0}$.

State and prove Claim 3, from earlier in Section 5.6, in the more general $\mu$-a.e.-on-$A$/$\mu$-measure-on-$A$ setting introduced in problem 5.6.10 (where each of $f, f_{1}, f_{2}, \ldots$ is only assumed defined and finite $\mu$-a.e.on $A$, rather than everywhere on $A$).

Claim 3 (the result to be generalized here): Let $(\Omega , \mathcal{F}, \mu )$ denote a measure space with $\mu (\Omega )<+\infty$, and let $\varnothing \subsetneq A \in \mathcal{F}$. Suppose that each of $f, f_{1}, f_{2}, \ldots$ is $\mathcal{F}/\mathfrak {B}^{*}$-measurable with domain $A$ and range contained in $\mathbb {R}$ such that $f_{n} \rightarrow f$ $\mu$-a.e.on $A$. Then $f_{n} \rightarrow f$ in $\mu$-measure on $A$.

State and prove Claim 4 (Egorov's Theorem), from earlier in Section 5.6, in the more general $\mu$-a.e.-on-$A$/$\mu$-measure-on-$A$ setting introduced in problem 5.6.10 (where each of $f, f_{1}, f_{2}, \ldots$ is only assumed defined and finite $\mu$-a.e.on $A$, rather than everywhere on $A$).

Claim 4 (Egorov's Theorem, the result to be generalized here): Let $(\Omega , \mathcal{F}, \mu )$ denote a measure space with $\mu (\Omega )<+\infty$, and let $A \in \mathcal{F}$ be nonempty. Let each of $f, f_{1}, f_{2}, \ldots$ be $\mathcal{F}/\mathfrak {B}^{*}$-measurable functions with domain $A$ and range contained in $\mathbb {R}$ such that $f_{n} \rightarrow f$ $\mu$-a.e.on $A$. Then for every $\epsilon >0$ there exists $B \in \mathcal{F}$ such that $\mu (B^{c})<\epsilon$ and $f_{n} \rightarrow f$ uniformly on $B$. (Roughly speaking, this says that if $f_{n} \rightarrow f$ $\mu$-a.e., then $f_{n} \rightarrow f$ uniformly except on a set of arbitrarily small $\mu$-measure — that is, $f_{n} \rightarrow f$ "uniformly almost everywhere.")