measure theory - $lim_{n to infty} int_X f_n , dmu = int_X f , dmu$ implies $lim_{n to infty} int_B f_n , dmu = int_B f , dmu$ for $B subseteq X$

I'm having trouble with the following problem.

Let $(X, \mathcal{M},\mu)$ be a measure space, where $X = [a,b] \subset \mathbb{R}$ is a closed and bounded interval and $\mu$ is the Lebesgue measure. Let $f_{n}$ be a sequence of non-negative functions in $L^{1}(X,\mathcal{M},\mu)$ $\textit{converging in measure}$ to a function $f \in L^{1}(X,\mathcal{M},\mu)$. Given that the following holds,

$\lim\limits_{n\rightarrow\infty}\int\limits_{X}f_{n}d\mu = \int\limits_{X}fd\mu$

show that for all $B \subset X$,

$\lim\limits_{n\rightarrow\infty}\int\limits_{B}f_{n}d\mu = \int\limits_{B}fd\mu$

where $B$ belongs to the Borel $\sigma$-algebra.

I was given a hint where convergence in measure in X implies convergence in measure in B, but I'm not sure where to proceed from here.