For a homework assignment I was told to prove that given $f\in L^1(\mathbb R)$, the following function is continuous $$F(x)=\int_{(-\infty,x]}fd\lambda.$$
I thought to use DCT and show sequential continuity. So given $x_n\rightarrow x$, we look at the sequence of indicators $\mathbf 1_{(-\infty,x_n]}f$. These are dominated by $|f|$ which is in $L^1$ iff $f$ is, which we know. So by DCT $$\lim _n \int \mathbf 1_{(-\infty,x_n]}f=\int \mathbf 1_{(-\infty,x]}f$$ as desired. Is my proof correct?
In fact, $F$ is uniformly continuous in $\mathbb R$.
To see this, let $\varepsilon>0$ and set $g_n(x)=\min\{\lvert\, f(x)\rvert,n\}$. By virtue of the Lebesgue Dominated Convergence Theorem, we have that $\|g_n-\lvert\, f\rvert\|_{L^1}\to 0$.
Fix $N>0$, such that $\|g_N-\lvert\, f\rvert\|_{L^1}<\varepsilon/2$. Then $$ \int_{\mathbb R}g_N\,dx\le \int_{\mathbb R}\lvert\, f\rvert\,dx\le\int_{\mathbb R}g_N\,dx+\frac{\varepsilon}{2}. $$
Set $\,\delta=\displaystyle\frac{\varepsilon}{2N+1}\,$ and let $\,\lvert x-y\rvert<\delta.\,$ Then $$ \lvert\,F(x)-F(y)\rvert=\left|\int_x^y f(t)\,dt\,\right|\le\int_x^y \lvert\,f(t)\rvert\,dt\le \frac{\varepsilon}{2}+\int_x^y g_N(t)\,dt \\ \le \frac{\varepsilon}{2}+N\lvert x-y\rvert \le \frac{\varepsilon}{2}+N\cdot\frac{\varepsilon}{2N+1}<\varepsilon. $$