How to verify this relationship between area under the graph and the preimage? Define $h : \mathbb{R} \to [0, \infty)$,
Let $H = \{(x,y)| 0 \leq y \leq h(x)\}$ be the area under the graph (including the boundary)

I wish to show the following is true:
$$H = \bigcup_{c>0} h^{-1}[c,\infty) \times [0, c]$$

Attempt: 
($\subseteq$) Let $(x,y) \in H$, then $0 \leq y \leq h(x)$, then $h^{-1}[y,\infty) \times [0, y] \subseteq h^{-1}[c,\infty) \times [0, c] \quad \forall y,c \in \mathbb{R}$, so $H \subseteq \bigcup_{c>0} h^{-1}[c,\infty) \times [0, c]$
($\supseteq$)  Not sure...

If $h = 0$ (so the constant function), then $h^{-1}[c,\infty)] = \emptyset$ for all $c > 0$, as no $x$ has $h(x) \ge c > 0$. And then the product set is empty too and we'd get $H = \emptyset$ which is false, as $H = X \times \{0\}$. So we should have a union for all $c \ge 0$ not just $c > 0$. The rest of the post assumes this change.
Now the proof from left to right should be written a bit differently. Pick $(x,y) \in H$, so $0 \le y \le h(x)$, where $x,y$ are fixed. Pick $c = y$, which is now OK, as $y \in [0,\infty)$, so $c \ge 0$. Then $x \in h^{-1}[[c,\infty)]$, as $h(x) \ge y = c$ and $y \in [0,c]$, as $c = y$. So $(x,y)$ is in the union, as we have a specific (!) $c$ such that $(x,y)$ is in the right hand side for that $c$.
Right to left is quite similar. If $(x,y)$ is in the union, there is a specific fixed $c$ (for that pair) such that $(x,y) \in h^{-1}[[c,\infty)] \times [0,c]$.
The latter means that $x \in h^{-1}[[c,\infty)]$ and $y \in [0,c]$. So we have
$h(x) \in [c,\infty)$ (so $h(x) \ge c$) and $0 \le y \le c$, and so combining we get $0 \le y \le c \le h(x)$ which means that $(x,y) \in H$. 
