Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions that converges in measure to $f$ on $[a,b]$. If there is a Lebesgue integrable function $g$ on $[a,b]$ such that $|f_n(x)|\leq g(x)$, $\forall n\geq 1$, $\forall x\in [a,b]$, then prove $\lim_{n\to\infty} \int_a^b f_n(x)dx=\int_a^b f(x)dx$.
I'm stuck on the "converges in measure" part. I don't know what I'm allowed to assume from that. And where does that $g$ fit in?
Any help would be greatly appreciated.
edit: Can I assume/show $[a,b]$ is a measurable set? Then wouldn't this just be Lebesgue's dominated convergence theorem?
edit2: Since $\{f_n\}_{n=1}^\infty$ is a sequence of measurable functions and $f$ is a measurable function and if I can assume/show that $[a,b]$ is a measurable set, can I say $f_n\to f$ for $n\to\infty$?
Idea: dominated convergence. It suffices to show that every convergent subsequence of $\int_a^b f_n(x)$ converges to $\int_a^b f(x)$
Suppose $\int_a^b f_{nk}(x)$ converges. Then since convergence in measure is assumed, {$f_{nk}$} contains a subsequence, {$f_{nkl}$} $\rightarrow $ f a.e. (Riesz Fischer)
By, the dominated Convergence Theorem, $\int_a^b f_{nkl}(x)$ $\rightarrow$$\int_a^b f(x)$
So, $\int_a^b f_{nk}(x)$ $\rightarrow$ $\int_a^b f(x)$
For all convergent $f_{nk}$ So,
$\int_a^b f_n(x)$ $\rightarrow$ $\int_a^b f(x)$
The End