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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?

Thanks in advance!!

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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.

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  • $\begingroup$ 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) $\endgroup$ – Riju Jul 7 '17 at 20:36
  • $\begingroup$ @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 $$(f_n+g_n)^2 \to (f+g)^2 \quad f_n^2 \to f^2 \quad g_n^2 \to g^2.$$ $\endgroup$ – saz Jul 8 '17 at 13:07
  • $\begingroup$ 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. $\endgroup$ – Riju Jul 8 '17 at 13:13

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