13
$\begingroup$

A sequence $\{f_n\}$ of measurable functions is said to be a Cauchy sequence in measure if, given $ϵ > 0$, there is an $N$ such that for all $m, n ≥ N$ we have $m\{x \in E : |f_n(x) − f_m(x)| \ge ϵ\} < ϵ$.

Show that if $\{f_n\}$ is Cauchy in measure, then there exists a measurable function $f$ to which the sequence $\{f_n\}$ converges in measure.

Idea: I need to show that $m\{x \in E : |f(x) − f_n(x)| \ge ϵ\} \rightarrow 0$ for some measurable function $f$. I was thinking that there exists $f(x) = \lim_{k\to \infty} f_{n_k}(x)$ where $f_{n_k}$ is a subsequence. But I am unclear as to where to proceed.

$\endgroup$
13
$\begingroup$

We essentially (as always) pass to a subsequence $\{g_n\}$ which is chosen such that if $E_j = \{x: |g_j(x) - g_{j+1}(x)| \geq 2^{-j}\}$ then $\mu(E_j) \leq 2^{-j}$. Let $F_k = \cup_{k}^\infty E_j$, so $\mu(F_k) \leq 2^{1-k}$ by subadditivity of $\mu$. For $x \notin F_k$ and $i \geq j \geq k$, you can show $$|g_j(x) - g_i(x)| \leq \sum_{l=j}^{i-1} |g_{l+1}(x) - g_l(x)| \leq 2^{1-j}$$ by definition of the subsequence.

So, $\{g_n\}$ is pointwise Cauchy on $F_k^C$. Let $F = \cap_1^\infty F_k$ (this is the limsup of the $E_j$'s), which has $\mu(F) = 0$. Then, on $F$, let $f=0$, and on $F^C$ let $f(x) = \lim_j g_j(x)$. Then, $g_j \to f $ a.e., and $|g_j(x) - f(x)| \leq 2^{1-j}$ for $x \in F_k^C$ and $j \geq k$. Then, note $\mu(F_k) \to 0$ and $g_j \to f$ in measure.

Finally, write $$\{|f_n(x) - f(x)|\geq \epsilon\} \subset \{|f_n(x) -g_j(x)| \geq 1/2 \epsilon\} \cup \{|g_j(x) -f(x)| \geq 1/2 \epsilon\}$$ and note that the right hand side can be made small with large $n$,$j$, and thus, you have convergence in measure.

This is taken from Theorem 2.30 in Folland's Real Analysis: Modern Techniques and their Applications 2e.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.