The Lebesgue integral of an integrable function is continuous

Let $a\in \mathbb{R}$ be fixed, and $f:\mathbb{R}\rightarrow \mathbb{R}$ a Lebesgue integrable function. Define $F:\mathbb{R}\rightarrow \mathbb{R}$ by $$F(x)=\int_a^xf(t)\mathrm{d}t$$ for all $t\in \mathbb{R}$. Prove that $F$ is continuous.

Using the definition of continuity involving convergent sequences, for $x_0\in \mathbb{R}$, I take any sequence $\lbrace x_n \rbrace$ such that $x_n \rightarrow x_0$ as $n\rightarrow \infty$. Then I need to prove that $F(x_n)\rightarrow F(x)$, i.e., $$\lim_{n\to \infty}\int_a^{x_n}f(t)\mathrm{d}t=\int_a^xf(t)\mathrm{d}t$$

I know that, if $m$ is the Lebesgue measure, then $$\lim_{n\to \infty}\int_a^{x_n}f(t)\mathrm{d}t=\lim_{n\to \infty}\int_{-\infty}^{\infty}\chi_{[a,x_n]}f\mathrm{d}m$$

Since $f$ is integrable by assumption, this implies $|f|$ is integrable as well. By the Dominated Convergence Theorem, $|\chi_{[a,x_n]}(t)f(t)|\leq |f(t)|$, so we interchange the limit and integral: $$\lim_{n\to \infty}\int_a^{x_n}f(t)\mathrm{d}t=\int_{-\infty}^{\infty}\lim_{n\to \infty}\chi_{[a,x_n]}f\mathrm{d}m$$

If I can show that $\lim_{n\to \infty}\chi_{[a,x_n]}(t)=\chi_{[a,x_0]}(t)$, then I am done, but this is not immediately obvious to me, nor is a proof of it.

If $x_n\to x_0$, then for any $a\leq t < x_0, \exists N\in \mathbb{N}$ such that $$t< x_n \quad\forall n>N$$ and hence $\chi_{[a,x_n]}(t) = 1$ for all $n>N$. Similarly, for $t>x_0, \exists M\in \mathbb{N}$ such that $\chi_{[a,x_n]}(t) = 0$ for all $n>M$. Since the singleton $\{x_0\}$ has measure zero, you can conclude that $$\lim \chi_{[a,x_n]} \to \chi_{[a,x_0]} \text{ a.e.}$$ and so Dominated convergence applies as you say.
Appeal to the Dominated Convergence Theorem. In fact you can weaken your hypothesis on $f$ to be $f\in\mathcal{L}^1_{\rm loc}(\mathbb{R})$. The function $f$ need only be integrable on bounded intervals.