Prove an interesting function $f$ is semicontinuos. 
Let $r_n$ be the sequence of all rational numbers and $$f(x)=\sum_{n,r_n<x}\frac{1}{2^n}.$$ Show that $f$ is lower semicontinuous in $R$.

Since $f$ is continuous on irrationals. 
It is also an increasing function. So we only need to look at limit from the left side to rationals actually satisfy the condition of lower semicontinuous.
Then I got stuck here. Could someone kindly help? Thanks!

Let $$ f_K (x) :=\sum_{n\leq K,\ r_n< x } \frac{1}{2^n}$$ Then
$f_K\rightarrow f $ uniformly since $\sum_{n>M} \frac{1}{2^n } <
\varepsilon$ for some $M$. Note that each $f_K$ is lower. So the
limit is lower.
(1) Lower Continuity of $f_K$ : Define $g_n$ :
$$
g_n(x) =\left\{
          \begin{array}{ll}
            \frac{1}{2^n}, & \hbox{$r_n<x$;} \\
            0, & \hbox{$x\leq r_n$;} \\
          \end{array}
        \right.
$$
is lower. And note that $f_K(x)=\sum_{i=1}^K g_i(x)$. So we have
claim that if we have two lower continuous $f,\ g$ then $f+g$ is
lower continuous : Assume that $x_n<x,\ x_n\rightarrow x$ Then $$
(f+g)(x_n)=f(x_n)+ g(x_n)\rightarrow f(x)+ g(x) =(f+g)(x)$$
(2) Uniform limit of lower continuous functions is lower : Assume that $x_k< x,\ x_k\rightarrow x $ Then $$ |f(x_k)-f(x)| \leq |f_K(x_k)-f(x_k) | +|f_K(x_k)-f_K(x) |+|f_K(x)-f(x) | $$
