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.
Best Answer
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.