Let $E$ be a measurable set with finite measure, $(f_n)$ be a sequence of real-valued measurable functions on $E$ and $f$ be a real valued measurable function on $E$. It is required to prove that if $(f_n)$ converges to $f$ in measure then $(f_n^2)$ converges to $f^2$ in measure. The following is my attempt.

Suppose $(f_n)$ converges to $f$ in measure. Let $(f_{n_k})$ be a subsequence of $(f_n)$. Then $(f_{n_k})$ converges to $f$ in measure and there exists a subsequence $(f_{{n_k}_r})$ that converges to $f$ pointwise a.e. on $E$. Hence $f^2$ is such that for any subsequence of the sequence $(f_n^2)$, there exists a further subsequence that converges to $f^2$ pointwise a.e. on $E$, and therefore $(f_n^2)$ converges to $f^2$ pointwise a.e. on E. Now since $m(E)$ is finite, we have $(f_n^2)$ converges to $f^2$ in measure.

Is the above argument alright? Thanks.

  • $\begingroup$ It's not true that a.e. convergence of a further subsequence $(f_{n_{k_{r}}})$ for each subsequence $(f_{n_{k}})$ to $f^2$ implies a.e. convergence of $f^2$. For an alternative solution: What do you know about convergence in measure? $\endgroup$ – amars Jun 25 '17 at 21:13
  • $\begingroup$ Why is it not true? I know it is true for sequences of real numbers but why not in this case? I've studied a chapter called "Modes of convergence". $\endgroup$ – Janitha357 Jun 25 '17 at 21:17
  • $\begingroup$ If this was the case, every $L^1$-convergent sequence $(f_n ) \rightarrow f$ would be a.e. convergent: Each subsequence $(f_{n_{k}})$ also converges to $f$ in $L^1$ and thus has a further subsequence that converges a.e. to $f$. A common counterexample is a sequence on $[ 0,1)$, where $f_1=1_{[ 0,\frac{1}{2})}$, $f_2=1_{[\frac{1}{2},1)}$, $f_3=1_{[ 0,\frac{1}{4})}$... This sequence converges to 0 in $L^1$ but there is no pointwise convergence in any point. $\endgroup$ – amars Jun 25 '17 at 21:33
  • $\begingroup$ I agree. Could you give a hint for an alternative solution? $\endgroup$ – Janitha357 Jun 25 '17 at 21:56

In Addition to the comment above, here is an alternative idea. There are probably more elegant solutions (e.g. more similar to your approach), but this should work as well.
\begin{align} &\mu \left(x\in E: \vert f^2(x)-f_n^2(x)\vert \geq \epsilon \right) \\ =&\mu \left(x\in E: \vert f(x)+f_n(x)\vert \vert f(x)-f_n(x)\vert \geq \epsilon \right)\\ =&\mu \left(x\in E: \vert f(x)+f_n(x)\vert \vert f(x)-f_n(x)\vert \geq \epsilon \text{ and } \vert f(x)+f_n(x)\vert > k \right) + \mu \left(x\in E: \vert f(x)+f_n(x)\vert \vert f(x)-f_n(x)\vert \geq \epsilon \text{ and } \vert f(x)+f_n(x)\vert\leq k \right)\\ \leq&\mu \left(x\in E: \vert f(x)+f_n(x)\vert > k \right) + \mu \left(x\in E: \vert f(x)-f_n(x)\vert \geq \frac{\epsilon}{k} \right)\\ \end{align}

By $\sigma$-continuity of $\mu$ and since $\mu (E) < \infty$, the first term converges to 0, for $k\rightarrow \infty$. For any fixed $k$, the second term converges to 0 for $n\rightarrow\infty$, since $f_n \rightarrow f$ in measure. Now, choosing sufficiently large $k$ and then letting $n\rightarrow \infty$ makes $$ \mu \left(x\in E: \vert f^2(x)-f_n^2(x)\vert \geq \epsilon \right) $$ arbitrarily small.

  • $\begingroup$ Can you give more details on why the first term converges to 0 when $\mu < \infty$? $\endgroup$ – Qinsheng Zhang Mar 17 at 19:18
  • $\begingroup$ The intersection of these sets for all k is the empty set, which has measure zero. All measures have what is called „continuity from above“, which means that the measures of such „decreasing“ sets converge to the measure of their intersection - if at least at some point the sets have finite measure. See here under „Basic Properties“: en.m.wikipedia.org/wiki/Measure_(mathematics) $\endgroup$ – amars Mar 17 at 20:04

Convergence in measure means that $\mu(\{x: |f_n(x) - f(x)| \ge \epsilon\})$ tends to $0$ as $n \to \infty$. Fix $\epsilon$ and $\eta > 0$, and choose sufficiently large $M, N > 0$ such that $\mu(\{x: |f(x)|> M\}) < \frac{\eta}{3}$ and $\mu(\{x: |f_n(x)|> M\}) < \frac{\eta}{3}$ for $n \ge N$.

(We can do this because $A_m = \{x: |f(x)| > m\}$ has $A_1 \supset ... \supset A_m \supset ...$ and the intersection $\cap_mA_m$ clearly is empty, so that since $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m) = 0$, at some point we must have $\mu(A_m) < \frac{\eta}{6}$. It's in this step that we have used the finite measure property, since otherwise $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m)$ isn't necessarily true. As to the $f_n$'s, for $N$ sufficiently large we have $|f_n(x)| < |f(x)| + \epsilon$ if $n \ge N$, except on a set $E_N$ with $\mu(E_N) < \frac{\eta}{6}$. To get our $M$, we take $m$ so large that $\mu(A_m) < \frac{\eta}{6}$ and thus $\mu(A_m \cup E_N) < \frac{\eta}{3}$. By selecting $M = m + \epsilon$ and $n \ge N$ we obtain $\{x: |f(x)|> M\} \subset A_m$ and $\{x: |f_n(x)|> M\} \subset A_m \cup E_N$, so we have our $M$ and $N$.)

Now $\{x: |f(x)|> M\} \cup \{x: |f_n(x)|> M\}$ has measure at most $\frac{2\eta}{3}$; on the complement of this set, we have \begin{eqnarray}|f_n(x)^2 - f(x)^2| &&=&& |f_n(x) - f(x)|\cdot|f_n(x) + f(x)| \le |f_n(x) - f(x)| \cdot \Big( |f_n(x)| + |f(x| \Big)\\ &&\le && |f_n(x) - f(x)|\cdot2M&& \end{eqnarray} and so by choosing $n$ large enough (and larger than $N$), we can make this expression less than $\epsilon$ on all of our remaining set, except for a part with measure less than $\frac{\eta}{3}$. Therefore $$\mu(\{x: |f_n(x)^2 - f(x)^2| \ge \epsilon\}) < \frac{2\eta}{3} + \frac{\eta}{3} = \eta$$ for $n$ large enough, where $\epsilon$ and $\eta$ were arbitrary - which proves the claim.

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    $\begingroup$ We don't know if $f(x)$ is finite for all $x$. $\endgroup$ – Coward Jul 23 '20 at 21:16

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