Convergence of an integral function and simple convergence of an indicator function Let $f \in L^2(\mathbb{R}_+)$ and set
\begin{equation}
\varphi(x) = \int_x^\infty e^{-\xi} f(\xi) d\xi
\end{equation}
a.e. $x \in \mathbb{R}_+$. I wish to prove that $\varphi$ is continous on $\mathbb{R}_+$, and in general, how to show that $\chi_{[x, \infty[}(\xi)$ converges to $\chi_{[y, \infty[}(\xi)$ when $x \rightarrow y$, since I thought about applying Lebegue's dominated convergence theorem and got stuck here.
Here $\chi_E$ is the indicator (characteristic) function of a set $E$. Thanks.
 A: Continuity of $\varphi$
Consider $x,y\in\mathbb{R}_+$ and w.l.o.g. let $y>x$. Then we have by applying Hölder's inequality
\begin{align}
 |\varphi(y)-\varphi(x)| \le \int_x^y e^{-\xi}|f(\xi)|~\text{d}\xi \le \|e^{-\cdot}\|_{L^2((x,y))}\|f\|_{L^2((x,y))}.
\end{align}
Since the $L^2$-norms are absolutely continuous w.r.t. the integration domain, the right hand side converges to 0 for $x\uparrow y$. Analogously you get the same result for $x\downarrow y$, which shows the continuity of $\varphi$. 
Pointwise convergence a.e. of the characteristic functions
Let be $z>y$. Then we have $\chi_{[y,\infty)}(z)=1$ and for since $x\rightarrow y$ we get for every $x$ with $|x-y|<|z-y|$ that $\chi_{[x,\infty)}(z)=1$. So, the characteristic functions converge pointwise everywhere on $(y,\infty)$. Similarly you can show that the converge pointwise everywhere on $(-\infty, y)$, too.
But they don't converge at $y$, too, because $\chi_{[y,\infty)}(y)=1$ but for every $x>y$, we have $\chi_{[x,\infty)}(y)=0$ and for every $x<y$ we have $\chi_{[x,\infty)}(y)=1$.
