Prove that if $(f_n)$ converges to $f$ in measure then $(f_n^2)$ converges to $f^2$ in measure.

Question

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.

Answer 1

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.

Answer 2

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.