# Doubts in proof of product of sequnce of convergence in measure in finite measure also converges in measure

Convergence in measure of products

I come across above question in which David Sir give proof of question .

But I am not able to understand I tried to understand form morning i think I got all except last line .

Actually I found above proof very constructive . Sorry But I am not able to intuition to prove theorem .

I really thankful if someone help me .

ANy help will be appreciated

Observe that $$|f_n-f|.|g_n|>2\delta^2 \implies |f_n-f|.|g|>\delta^2 \ or \ |f_n-f|.|g_n-g|>2\delta^2$$ by triangle inequality and hence we get

$$\mu(\{|f_n-f|.|g_n|>2\delta^2\}\cap B)\leq\mu(\{|f_n-f|.|g_n-g|>\delta^2\}\cap B)+\mu(\{|f_n-f|.|g|>\delta^2\}\cap B)$$ $$\implies \mu(\{|f_n-f|.|g_n|>2\delta^2\}\cap B)\leq \mu(\{|f_n-f|.|g_n-g|>\delta^2\})+\mu(\{|f_n-f|.|g|>\delta^2\}\cap B)$$

where in the last inequality we have ignored the measurable set $$B$$ from the first summand.

Now $$|f_n-f|.|g_n-g|>\delta^2 \implies |f_n-f|>\delta \ or \ |g_n-g|>\delta$$ and hence we have $$\mu(\{|f_n-f|.|g_n-g|>\delta^2\})\leq\mu(|f_n-f|>\delta)+\mu(|g_n-g|>\delta)$$

For the second term we have $$\{|f_n-f|.|g|>\delta^2\}\cap B \subset \{|f_n-f|.|g|>\delta^2\}\cap \{|g|\frac{\delta^2}{A} \}$$. Thus we have $$\mu(\{|f_n-f|.|g|>\delta^2\}\cap B)\leq\mu(\{|f_n-f|>\frac{\delta^2}{A} \})$$

Plugging all these back in our original inequality we get

$$\mu(\{|f_n-f|.|g_n|>2\delta^2\}\cap B)\leq \mu(\{|f_n-f|.|g_n-g|>\delta^2\})+\mu(\{|f_n-f|.|g|>\delta^2\}\cap B)$$ $$\leq\mu(|f_n-f|>\delta)+\mu(|g_n-g|>\delta)+\mu(\{|f_n-f|>\frac{\delta^2}{A} \})$$.

This solves your problem because each summand individually goes to $$0$$ as $$n\rightarrow \infty$$