How do I show that this function is increasing, and has discontinuities at points in this sequence I have a series $\sum c_n$ that converges, and $c_n>0$. Then $(x_n)$ is a sequence in $(0,1)$. Let $f(x)=\sum\limits_{n=1}^\infty c_nf_n(x)$ where 
$$f_n(x) = 
     \begin{cases}
       1 & : x \geq x_n\\
       0 & : x < x_n
     \end{cases}
   $$
I've shown that $\sum\limits_{n=1}^\infty c_nf_n(x)$ indeed converges uniformly on $(0,1)$, and I've shown that $f$ is continuous for each $x\neq x_n$. Now I need to prove that $f$ is increasing on $(0,1)$, and that it's discontinuous for all $x=x_n$.
My idea for showing $f$ is increasing is to let $a<b$, then $f(b)-f(a)=\sum\limits_{n=1}^\infty c_n(f_n(b)-f_n(a))$. Then $f_n(b)-f_n(a)$ is either $1$ or $0$, and since $c_n>0$, $f(b)-f(a)\geq 0$. Is this right?
Then, how can I conclude that $f$ is not continuous when $x=x_n$? 
 A: Note that 
$$
f = \sum\limits_{n=1}^\infty c_n \chi_{[x_n,\infty)},
$$
basically a staircase function which is zero to the left of $0$, and jumps by $c_n$ at every $x_n$, and eventually stabilizes at the value $\sum c_n$ to the right of $0$.
Written this way, it should be clear that the larger $x$ is, the more strictly positive terms will be included in the sum defining $f(x)$.
Now let us show that $f$ is discontinuous at $x = x_n$. For this it is enough to show that it is not left-continuous. What will happen if you take a sequence $y_j$ such that $y_j<x$ for all $j$, and $y_j \rightarrow x$? Well, $f(y_j)$ will be missing that "$c_n$" term since $\chi_{[x_n,\infty)}(y_j) = 0 $ for all $j$. This means that $f(y_j)$ will always differ from $f(x)$ by at least $c_n >0$.
By the way, this can be used to show that there is a continuous monotonic function which is discontinuous precisely at the rationals. It is an interesting consequence of the Baire category theorem that there is no continuous function which is discontinuous precisely at the irrationals.
