Measure theory convergence in measure

Question

If $f_n(x) $ be a sequence of measurable and finite almost everywhere functions on $[a, b] $ that is converging in measure to $f(x)$ finite almost everywhere and measurable on [a, b]. If $ g(x) $is finite almost everywhere function and measurable on$ [a, b]$ then $g(x)f_n(x) $ converging to $f(x)g(x)$ in measure. How can i prove that? Any hint?

Answer 1

Let $N$ be a set of measure zero such that $|g(x)|<\infty$ for all $x\in[a,b]-N$.

Assume that $f_{n}\rightarrow f$ in measure. Now fix a subsequence $(f_{n_{k}})$ of $(f_{n})$. As $f_{n_{k}}\rightarrow f$ in measure. Then there exists a further subsequence $(f_{n_{k_{l}}})$ such that $f_{n_{k_{l}}}(x)\rightarrow f(x)$ a.e.

Without loss of generality, we can assume that $f_{n_{k_{l}}}(x)\rightarrow f(x)$ for all $x\in[a,b]-N$. Then so is $f_{n_{k_{l}}}(x)g(x)\rightarrow f(x)g(x)$ for all such $x$.

This proves that $f_{n}g\rightarrow fg$ in measure.

The theorem used here is that, in a finite measure space, a sequence $(f_{n})$ converges to $f$ in measure if and only if for every subsequence $(f_{n_{k}})$ of $(f_{n})$, there exists a further subsequence $(f_{n_{k_{l}}})$ such that $f_{n_{k_{l}}}(x)\rightarrow f(x)$ a.e.