Superior Riemann integral identity as the "volume" of a set

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I'm trying to solve the following problem,

Let $f: A \subset \mathbb{R}^m \to \mathbb{R}$ such that $0 \leq f(x) \leq M$, for all x in the rectangle A. Consider the set $$C(f) = \{ (x,y) \in A \times [0,M] \, | \, 0 \leq y \leq f(x) \}$$. Show that, $$\int_{A}^{-} f(x) dx = \int_{A \times [0,M]}^{-} \chi_{C(f)}(x) dx, $$ where $\chi_{C(f)}(x)$ denotes the characteristic function of the set $C(f)$.

Intuitively, the problem is saying that the superior Riemann integral of a function is in some way the "volume below the graph of f".

To attack this problem, I have tried to show the reciprocal inequalities, i.e., $\int_{A}^{-} f(x) dx \leq \int_{A \times [0,M]}^{-} \chi_{C(f)}(x) dx$, and, $\int_{A}^{-} f(x) dx \geq \int_{A \times [0,M]}^{-} \chi_{C(f)}(x) dx.$

But I don't get arrive nowhere.

Any hints?