Semicontinuous and monotone = Semicontinuous? Is it true for a (upper/lower) semicontinuous function $f$ and a monotonically increasing function $g$ which is also (upper/lower) semicontinuous) that $f\circ g$ is also (upper/lower)semicontinuous?
Somehow I think that this is true( just a Feeling), but I don't know which equivalent definition of semicontinuity I can use to prove this?
 A: Claim 1. If $f$ and $g$ are lower semicontinuous and $g$ is monotone then $f\circ g$ can fail to be lower semicontinuous.
Indeed, take $f(x)=-x$ and $g(x)=\begin{cases}x & x\le 0 \\ x+1 & x>0\end{cases}$. Then $f\circ g$ is not lower semicontinuous.
Note. In this example $f\circ g$ is however upper semicontinuous. But one can make $f\circ g$ worse by taking more complicated $f$, for instance $f(x)=\begin{cases}-x & x \le 1 \\ \sin\frac{1}{x-1} & x>1\end{cases}$.
In this case $f\circ g$ is neither lower nor upper semicontinuous. (Observe that the graph of $f\circ g$ is obtained from the graph of $f$ essentially by "cutting" away $(0,1]$ from $x$-axis...)
Claim 2. If $f$ and $g$ are lower semicontinuous and $g$ is monotone then $g\circ f$ is lower semicontinuous.
Let us fix $x_0$.
By lower semicontinuity of $g$ for any $\varepsilon>0$ there exists $\delta>0$ such that $g(f(x_0)-\delta) \ge g(f(x_0))$
Now, by lower semicontinuity of $f$ there exists $\gamma >0$ such that $f(x)\ge f(x_0) - \delta$ if $|x-x_0|<\gamma$. Using monotonicity of $g$ we get $g(f(x)) \ge g(f(x_0)- \delta)$. And using the inequality we obtained earlier we get $g(f(x)) \ge g(f(x_0))$, hence $g\circ f$ is lower semicontinuous.
Note. If $g$ is just monotone, Claim 2 does not hold. Just take any monotone not semicontinuous function $g$ and $f(x)=x$. (Sorry if my comments were confusing)
