# If $f_{n}\to f$ in measure, then $f_{n} \cdot g \to f \cdot g$ in measure

This question has been asked in SE before, but my question asks to prove in a different way.

Let $E$ be a set of finite lebesgue measure. $\{\ f_{n} \}\ \to f$ in measure on $E$, and $g$ be a measurable function which is finite a.e on $E$. Prove that $f_{n}.g \to f.g$ in measure. Deduce from that $\{\ f_{n}^{2} \}\ \to f^{2}$in measure. Infer from this that if $\{\ g_{n} \}\ \to g$ in measure on $E$, then $\{\ f_{n} \cdot g_{n} \}\ \to f \cdot g$ in measure on $E$.

I have shown the first part. But I am struggling to show that $\{\ f_{n}^{2} \}\ \to f^{2}$ in measure. What I have is there exists a subsequence $f_{n_{k}}$ of $f$, such that $f_{n_{k}} \to f$ a.e on $E$. Now we have $f_{n_{k}}^{2} \to f^{2}$ a.e on $E$. Then since $E$ is of finite measure we have $f_{n_{k}}^{2} \to f^{2}$ in measure on $E$. But how to get $f_{n}^{2} \to f$ in measure on $E$.

For the last part it seems that I have to prove that any linear combination of sequences converging in measure also converge in measure. Then I have to use that $$2f_{n} \cdot g_{n}=(f_{n}+g_{n})^{2}-f_{n}^{2}-g_{n}^{2}$$

Is that right? What to do about the second part and how to use the first part in the second part?

You can apply the subsequence principle to finish the proof of the first assertion:

Subsequence principle: Let $(a_n)$ be a sequence. Then $a_n$ converges if, and only if, there exists an element $a$ such that for any subsequence $(a_{n_k})_k$ there exists a subsequence $(a_{n_{k_{\ell}}})_{\ell}$ such that $a_{n_{k_{\ell}}} \to a$.

Since your reasoning shows that for any subsequence $(f_{n_k})_k$ there exists a subsequence $(f_{n_{k_{\ell}}})$ such that $f_{n_{k_{\ell}}}^2 \to f^2$ in measure, we get $f_n^2 \to f^2$ in measure.

Regarding the second part: Yes, using

$$2 f_n g_n = (f_n+g_n)^2 -f_n^2 - g_n^2$$

is a good idea. Just note that, by the first part, $(f_n+g_n)^2 \to (f+g)^2$, $f_n^2 \to f^2$ and $g_n^2 \to g^2$ in measure.

• Yes this is the solution I already have. But what I want to know is how this part(1) is helping me to prove part (2) – Riju Jul 7 '17 at 20:36
• @Riju Are we talking about different things...? Part 1: Show that $f_n \to f$ implies $f_n^2 \to f^2$. Part 2: $f_n \to f$, $g_n \to g$ implies $f_n g_n \to f g$. As I pointed out in my answer you need part 1 to show part 2 since we need to show that $$(f_n+g_n)^2 \to (f+g)^2 \quad f_n^2 \to f^2 \quad g_n^2 \to g^2.$$ – saz Jul 8 '17 at 13:07
• am really sorry, I have edited the question now. Actually the first part is $f_{n}.g \to f.g$ in measure. This is what I have already proved. Now I want to prove $f_{n}^{2} \to f^{2}$ using that result. – Riju Jul 8 '17 at 13:13