How to prove this function is integrable?? Let $f(x)=0$ when $x<0$, and $f(x)=1$ if $x\geq 0$. Choose a countable dense sequence $\{r_n\}$ in [0,1]. Then, show that the function $F(x)=\sum_{n=1}^\infty  1/n^2 f(x-r_n)$ is integrable and has discontinuities at all points of the sequence $\{r_n\}$.
I will prove this problem by using thm "if {x; f is discontinuous} has measure zero and f is bounded, then f is integrable". 
But, I cannot prove that F is discontinuous at seq $\{r_n\}$. 
How to prove that?
 A: If $x = r_k \in (0,1)$ for some $k$, then for all $y < x$ we have
\begin{align}
f(x) - f(y) &= \sum_{n=1}^\infty \frac{f(x-r_n)}{n^2} - \sum_{n=1}^\infty \frac{f(y-r_n)}{n^2} \\
&= \sum_{n=1, r_n \leq x}^\infty \frac{1}{n^2} - \sum_{n=1, r_n \leq y}^\infty \frac{1}{n^2} \\
& \geq \frac{1}{k^2}\,,
\end{align}
since thes sum for $f(x)$ contains all the terms that the sum for $f(y)$ contains, and the sum for $f(x)$ contains the term $1/k^2$, which is not contained in the sum for $f(y)$ because $y < x = r_k$. Thus $f$ is not continuous at the points $\{r_j\}$.
Now if $x$ is not a point of the sequence $\{r_j\}$, then for all $n$ there exists a $\delta$ such that the interval $(x- \delta, x + \delta)$ does not contain the points $\{r_1, ..., r_n\}$. Now for $y \in (x- \delta, x)$ we have
\begin{align}
|f(x) - f(y)| &= \sum_{j=1, r_j \leq x}^\infty \frac{1}{j^2} - \sum_{j=1, r_j \leq y}^\infty \frac{1}{j^2} \\
&= \sum_{j=1,y \leq r_j \leq x} \frac{1}{j^2} \\
& \leq \sum_{j=n+1}^\infty \frac{1}{j^2}\,.
\end{align}
The last inequality follows from the fact that $(x- \delta, x)$ does not contain any of the points $\{r_1, ..., r_n\}$. As a residue of a convergent series this tends to $0$ as $n \to \infty$. Thus $f$ is left-continuous at $x$. Similar computation shows right-continuity.
