# Prove that if $f_n \rightarrow f$ in measure and $\phi$ is continous, then $\phi \circ f_n \rightarrow \phi \circ f$ in measure.

Let $$f_n,f: E \rightarrow \mathbb{R}$$ where $$E$$ has finite measure, suppose that $$\phi:\mathbb{R} \rightarrow \mathbb{R}$$ is continuous. Then $$\phi \circ f_n \rightarrow \phi \circ f$$ in measure. Show that this can fail when $$E$$ has infinite measure.

Here is my work so far: Since $$f_n \rightarrow f$$ in measure there exists a subsequence $$f_{n_k}$$ such that $$f_{n_k} \rightarrow f$$ pointwise a.e. on $$E$$. Then by continuity we have $$\phi\circ f_{n_k} \rightarrow \phi \circ f$$ pointwise a.e. on E. Of course this implies that $$\phi \circ f_{n_k} \rightarrow \phi \circ f$$ in measure. My trouble is that this only shows a subsequence converges in measure and not the full sequence itself. I feel like I almost there but I am just confused about how to end the proof.

Any help is appreciated!

• Go down to the level of sub-subsequences. Mar 17, 2021 at 18:24
• So consider a subsequence of $f_n$, say $f_{n_k}$. Since $f_{n_k}$ converges to $f$ in measure we do the same as before, find a sub-subsequence that converges to $f$ pointwise a.e. This implies that $\phi \circ f_{n_{k_j}} \rightarrow \phi \circ f$ in measure. How do we continue from here? Mar 17, 2021 at 18:39
• @Blaze are you familiar with the characterization that $f_n \to f$ in measure (finite measure space) if and only if for every subsequence of $f_n$ there is a further subsequence that converges a.e.? Mar 17, 2021 at 18:40
• @nullUser I am not familiar with this, but I can see how it would solve my problem immediately. Mar 17, 2021 at 18:42

Theorem. Let $$\mu$$ be a finite measure. Then $$(f_n)$$ converges to $$f$$ in measure if and only if for every subsequence of $$(f_n)$$ there is a further subsequence that converges almost everywhere.
Proof. Suppose $$(f_n)$$ converges to $$f$$ in measure, i.e. $$\mu(|f_n - f| > \epsilon) \to 0$$ for all $$\epsilon > 0$$, and let a subsequence be given. For convenience just write it as $$(f_n)$$ again to avoid subscripts. Take a sequence $$\epsilon_n \to 0$$ decreasing and $$\delta_n$$ decreasing with $$\sum_n \delta_n < \infty$$ and choose $$N(0) = 0$$ and $$N(n) > N(n-1)$$ such that $$\mu(|f_{N(n)} - f| > \epsilon_n) < \delta_n$$ for all $$i \geq N(n)$$. Since $$\sum_n \delta_n <\infty$$, by the Borel-Cantelli lemma, except on a set of measure $$0$$, one has $$|f_{N(n)}-f| \leq \epsilon_n$$ for all but finitely many $$n$$. Since the $$N(n)$$ are increasing and $$\epsilon_n$$ decreasing, $$|f_{N(i)}-f| \leq \epsilon_i \leq \epsilon_n$$ for all $$i \geq n$$ except finitely many exceptions. Thus $$f_{N(n)} \to f$$ as $$n \to \infty$$ except on a set of $$0$$ measure.
Next suppose that for each subsequence there is a further subsequence converging almost surely. Assume for contradiction that $$\limsup_n \mu(|f_n - f| > \epsilon) = \delta > 0$$ for all $$n$$. Then choose a subsequence (which to avoid subscripts we assume is the whole sequence) such that $$\mu(|f_n - f| > \epsilon) \geq \delta/2$$ for all $$n$$. By the assumed property, choose a further subsequence such that $$f_{N(n)} \to f$$ except on a set of measure $$0$$. Then $$\delta/2 \leq \mu(|f_{N(n)}-f| > \epsilon) = \int 1_{|f_{N(n)}-f|>\epsilon} d\mu \to 0$$ by the dominated convergence theorem (using here that $$\mu$$ is finite), a contradiction. Thus it must be that $$f_n \to f$$ in measure.