It depends on what you can assume.
If you know Lebesgue integration then it is easy consequence of Dominated Convergence Theorem.
If you do not know Lebesgue Theory then you can prove DCT directly without "measure things". Of course it is much more than you ask but also much more general and with quite simple proof.
Theorem. Let $(f_n)$ be a sequence of Riemann integrable functions on $[0,1]$ such that $|f_n(x)| < M$ for $x \in [0,1]$ and $f_n(x) \to f(x)$ on $[0,1]$. Then $ \lim_{n \to \infty} \int_0^1 f_n(x) dx = \int_0^1 \lim_{n \to \infty} f_n(x) dx. $
Proof. By linearity of integral we can assume that $f \equiv 0$ and $f_n(x) > 0$. Then we need to show that $\int_0^1 f_n(x) dx \to 0$ as $n \to \infty$. Assume that it is not true.
Let $F_n := f_n/M$. Then $|F_n(x)| < 1$, $F_n(x) \to 0$ for $n \to \infty$ and by the assumption $I := \inf_n \int_0^1 F_n(x) dx > 0$ (here we choose suitable subsequence of $(F_n)$).
Let $s_n$ be a lower Riemann sum of $F_n$ such that $s_n \geq \frac{3}{4}I$. Than if $u_n$ is the sum of lengths of intervals of the partition of $[0,1]$ such that $F_n(x) > \frac{1}{2} I$ (on that intervals) then by $F_n(x) < 1$ and the definition of a lower sum we obtain $\frac{3}{4} I \leq s_n \leq u_n + \frac{1}{2} I (1 - u_n) = \frac{1}{2} I + u_n (1-\frac{1}{2} I). $ Hence $u_n \geq \frac{1}{4 I (1-I/2)} > 0$.
Let $U_n$ be the set of interiors of intervals defined by $u_n$. Define $V_n = \bigcup_{k=n}^\infty U_n$. Then each $V_n$ is open and the sequence $(V_n)$ is decreasing with $\inf_n |V_n| > 0$. Hence $\bigcap_n V_n \neq \emptyset$ and there exists $x_0$ such that $x_0 \in \bigcap_n V_n$. Then $F_n(x_0) > \frac{1}{2} I$ thus $F_n(x_0) \not\to 0$ and we obtain a contradiction.