Problem with approximation of semicontinuous function with continuous functions Assume that a function  $f: \mathbb R \rightarrow \mathbb R$  is a lower semicontinuous. 
The Baire's theorem says that there is an increasing  sequence of continuous functions $(f_n)$ which is pointwise convergent to $f$. 
I known a proof in case when $f$ is additionally bounded. But I don't know how to proof this theorem for unbounded $f$. I try taking $h(x)=\frac{2}{\pi} \arctan f$ and next taking an increasing  sequence of continuous functions $(g_n)$ pointwise convergent to $h$. But it may happens that $g_n(x)<-1$ for some $x$ and I cannot take $f_n(x)=\tan (\frac{\pi}{2}g_n(x))$. 
How to prove the Baire theorem for unbounded $f$?
 A: I suppose the problem is with a lower semicontinuous function that is not bounded below. If the function is bounded below (without loss of generality $f \geqslant 0$) but not above, one can approximate the bounded function $\tilde{f}(x) = \dfrac{f(x)}{1+f(x)}$ by an increasing sequence $g_n$, cap each $g_n$ at $1-\frac1n$, $h_n(x) = \min \left\{ g_n(x),\, \frac{n-1}{n}\right\}$, and use
$$k_n(x) = \frac{h_n(x)}{1-h_n(x)}$$
for the approximating monotonic sequence of continuous functions.
For $f$ not bounded below, if we find a continuous $g \leqslant f$, we can reduce the approximation to the above, $h = f-g \geqslant 0$ is lower semicontinuous, and if $k_n \uparrow f-g$, then $k_n+g \uparrow f$.
So it remains to find a (finitely valued) continuous $g \leqslant f$. Since a lower semicontinuous function attains its minimum on any compact subset of $\mathbb{R}$, we have no problem finding a continuous function $g_{a,b}$ on $[a,b]$ that is a lower bound of $f$ there (for example a constant function). Then, using a partition of unity, we can glue those lower bounds together to obtain a global continuous $g\leqslant f$.
For example, let
$$\psi(x) = \begin{cases}\qquad 0 &, \lvert x\rvert \geqslant \frac34\\
2\left(x+\frac34\right) &, -\frac34 \leqslant x \leqslant -\frac14\\
\qquad 1 &, \lvert x\rvert \leqslant \frac14\\
2\left(\frac34-x\right) &, \frac14 \leqslant x \leqslant \frac34 \end{cases}$$
Then the translates of $\psi$ by integers form a continuous partition of unity,
$$\sum_{k\in\mathbb{Z}} \psi(x-k) \equiv 1,$$
and letting
$$c_k := \min \{ f(x) : x \in [k-1,\,k+1]\}$$
we obtain a continuous function $g \leqslant f$ by setting
$$g(x) = \sum_{k\in\mathbb{Z}} c_k\cdot \psi(x-k).$$
