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 enter image description here

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


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