Let $f$ be integrable on $\mathbb{R}$ and for $t \in \mathbb{R}$ define the function $F$ by $F(t) = \int_{(-\infty,t]} f(x)dx$. Prove that $F$ is continuous, uniformly bounded and that:
$lim_{t \rightarrow -\infty} F(t) = 0$
and that
$lim_{t \rightarrow \infty} F(t) = \int_{\mathbb{R}}fdx$
So this result makes a lot of sense. Since $f$ is integrable $F$ should converge to the $\int_{\mathbb{R}}f$ as $t \rightarrow \infty$, and converge to $0$ as $t \rightarrow -\infty$.
So, I can easily show that, $F$ is uniformly bounded by the value of $\int_\mathbb{R} f$. Can somebody help me to show that $F$ is continuous? Rigorous details appreciated!
The convergences as $t$ goes to $-\infty,\infty$ also make sense, but I was hoping somebody could give me insight as to how to prove it!! Again, rigorous details appreciated! Thanks!
HINT
Use sequential characterization of limits.
Let $t_n \to +\infty$
Denote $f_n=f(x)1_{(-\infty,t_n]}$
Then $f_n \to f(x)$ and by Dominated convergence you have the conclusion.
Do the same for an arbitrary sequence $y_n \to -\infty$