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
 A: 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 $\{|f_n-f|.|g|>\delta^2\}\cap B \subset \{|f_n-f|.|g|>\delta^2\}\cap \{|g|<A \} \subset \{|f_n-f|>\frac{\delta^2}{A} \}$. 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 $n\rightarrow \infty$
