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!

  • 3
    $\begingroup$ Go down to the level of sub-subsequences. $\endgroup$ Mar 17, 2021 at 18:24
  • $\begingroup$ 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? $\endgroup$ Mar 17, 2021 at 18:39
  • 1
    $\begingroup$ @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.? $\endgroup$
    – nullUser
    Mar 17, 2021 at 18:40
  • $\begingroup$ @nullUser I am not familiar with this, but I can see how it would solve my problem immediately. $\endgroup$ Mar 17, 2021 at 18:42

1 Answer 1


As mentioned in the comments, the following theorem trivializes the problem.

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.


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