Show that the function $h$ defined by $h(x)=g(f(x))$ is lower semicontinuous. Let $f:\mathbb{R}^n\longrightarrow \mathbb{R}$ be a lower semicontinuous function, and $g:\mathbb{R}\longrightarrow\mathbb{R}$ be a lower semicontinuous and nondecreasing function.
(1) Show that the function $h$ defined by $h(x)=g(f(x))$ is lower semicontinuous.
(2) Give an example showing that the nondecrease assumption is essential.

Part (1) 
$f$ is lower semicontinuous, therefore: $\forall x_0\in\mathbb{R}^n$
$$\forall\epsilon>0: \exists U\in N(x_0): f(x)\geq f(x_0)-\epsilon,\; \forall x\in U \tag{1}$$
Where $U$ is a neighborhood of $x_0$.
$g$ is lower semicontinuous, therefore: $\forall y_0\in\mathbb{R}$
$$\forall\epsilon>0: \exists V\in N(y_0): g(y)\geq g(y_0)-\epsilon,\; \forall y\in V \tag{2}$$
It is sufficient to prove for $h(x)=g(f(x))$ that: $\forall x_0\in\mathbb{R}^n$
$$\forall\epsilon>0: \exists U\in N(x_0): g(f(x))\geq g(f(x_0))-\epsilon,\; \forall x\in U \tag{3}$$

Attempt 
$g$ is nondecreasing, therefore $(1)$ becomes: $\forall x_0\in\mathbb{R}^n$
$$\forall\epsilon>0: \exists U\in N(x_0): g(f(x))\geq g(f(x_0)-\epsilon),\; \forall x\in U \tag{4}$$
I am stuck here. I am also looking at the possibility of taking $y_0=f(x_0)$ and $y=f(x)$.
I am frankly lost in the abstraction, and I do not fully understand why the lower semicontinuity is important here, or why $g$ must be nondecreasing.

Part (2)
I feel if I understand Part (1) then it would be easier for me to come up with an example.

Thank you.
 A: It might help to work with the sequential characterisation of lsc. A function $h$ is lsc.
at $x$ iff $\liminf_n h(x_n) \ge h(x^*)$ for any sequence $x_n \to x^*$.
Pick some sequence $x_n \to x^*$. Since $f$ is lsc. we have $\liminf_n f(x_n) \ge f(x^*)$, and since $g$ is
non decreasing we have $g(\liminf_n f(x_n) ) \ge g(f(x^*))$.
We have
$\lim_n \inf_{k \ge n} f(x_k) = \liminf_n f(x_n) $ (the definition) and so
$\liminf_n g(\inf_{k \ge n} f(x_k)) \ge g(\liminf_n f(x_n)) \ge g(f(x^*))$.
Since $g$ is non decreasing and 
$f(x_n) \ge \inf_{k \ge n} f(x_k)$ we have $g(f(x_n)) \ge g(\inf_{k \ge n} f(x_k))$, and
so $\liminf_n g(f(x_n)) \ge \liminf_n g(\inf_{k \ge n} f(x_k)) \ge g(f(x^*))$ and
so $g \circ f$ is lsc.
As an example of why the non decreasing is necessary, we can take a continuous
$g(x) = -x$ and let $f = 1_{(0,\infty)}$ (indicator function of the strictly positive
reals). Then with $x_n = {1 \over n}$ we have $f(x_n) = 1$ and
$\liminf_n f(x_n) = 1 \ge f(0) = 0$, but $g(f(x_n)) = -1$ and $\liminf_n g(f(x_n)) = -1$,
but $g(f(0)) = 0 > -1$,
so $g \circ f $ is not lsc.
