Pointwise and Measure Limits

Why

How does convergence pointwise (or almost everywhere pointwise) relate to convergence in measure?

Results

There exists a measure space and a sequence of measurable real-valued functions on that space converging everywhere (and so almost everywhere) but not converging in measure.

There exists a measure space and a sequence of measurable real-valued functions on that space converging in measure but not converging almost everywhere (nor everywhere).

On finite measure spaces, all sequences of measurable real-valued functions converging almost everywhere converge in measure.

Let $(X, \mathcal{A} , \mu )$ be a measure space. Let $\seq{f}$ be a sequence of measurable functions on $X$ such that $\seqt{f} \goesto f$ almost everywhere. Let $\epsilon > 0$.

For each $x \in X$, if $\abs{f_n(x) - f(x)} > \epsilon $ for infinitely many $n$, then $f_n(x) \not\goesto f(x)$. Let $A$ be the set of such $x$ and let $B = \Set*{ x \in X }{ f_n(x) \not\goesto f(x) }$. $A$ is a subset of $B$. The measure of $B$ is zero since $f_n \goesto f$. Use the the monotonicity of measure to conclude. $\mu (A) \leq \mu (B) = 0$. Since $\mu (A) \geq 0$, $\mu (A) = 0$.

For natural $k$, let $E_k$ be the $\Set*{x \in X} {\abs{f_k(x) - f(x)} > \epsilon }$. Then $x \in A$ means that for every natural $n$, there exists a $k \geq n$ such that $x \in E_k$. In particular, for every $n$, $x$ is in $\cup_{k = n}^{\infty}E_k$; denote this set by $B_n$. If $x$ is in $B_n$ for every $n$, then $x \in \cap _{n = 1}^{\infty} B_n$. So we can write

\[ A = \bigcap_{n = 1}^{\infty} \bigcup_{k = n}^{\infty} E_k = \bigcap_{n = 1}^{\infty} B_n . \]

The sequence of sets $\seq{B}$ is decreasing. So since $\mu $ is finite,

\[ \lim_{n \to \infty} \mu (B_n) = \mu (A) = 0. \]

For every $n$, the set $B_n$ contains $\Set*{x \in X}{\abs{f_n(x) - f(x)} > \epsilon }$, namely $E_n$, the first set in the union. So then $\mu (E_n) \leq \mu (B_n)$ by monotonicity and so

\[ 0 \leq \lim_{n \to \infty} \mu (E_n) \leq \lim_{n \to \infty} \mu (B_n) = 0, \]

and we conclude $\lim_{n} E_n = 0$. Since $\epsilon $ was arbitrary, we conclude $f_n \goesto f$ in measure.

On any measure space, for a sequence of measurable real-valued functions converging in measure to a measurable real-valued limit function, there exists a subsequence convergeng to the limit function almost everywhere.

Let $(X, \mathcal{A} , \mu )$ be a measure space. Let $\seq{f}$ be a sequence of measurable functions on $X$ such that $\seqt{f} \goesto f$ in measure.

There exists $n_1$ so that

\[ \mu (\Set*{x \in X}{\abs{f_{n_1}(x) - f(x)} > 1}) < \frac{1}{2}. \]

Can find $n_2 > n_1$ so that

\[ \mu (\Set*{x \in X}{\abs{f_{n_2}(x) - f(x)} > \frac{1}{2}}) < \frac{1}{4}. \]

We can inductively find a sequence $\set{n_k}_k$ so that:

\[ \mu (\Set*{ x \in x }{ \abs{f_{n_k}(x) - f(x)} > \frac{1}{k} }) \leq \frac{1}{2^k}. \]