Integrate joint pdf on a set of measure zero

Question

A set $A$ of $(x,y)$ values for which $\int \int_{A} dxdy=0$. Intuitively, if we integrate the joint pdf $f(x,y)$ on the set $A$, we will have zero. So any ideas on how to get this rigorously? I think that if $f(x,y)$ is bounded it is easy since suppose $$N \leq f(x,y) \leq M$$ then $$0=N\int \int_{A} dxdy=\int \int_{A}N dx dy \leq \int \int_{A}f(x,y)dxdy \leq \int\int_{A}Mdxdy=M\int \int_{A}dxdy=0$$ But we do not always have the bouded pdfs. So is there a simple way to prove this?

Answer 1

Note that density functions are nonnegative.

Let $S_r := \{(x,y) : f(x,y) > r\}$ and let $S_r^c$ be the complement. Then $$ \begin{align} 0 \le \iint_A f(x,y) \, dx \, dy &= \iint_{A \cap S_r^c} f(x,y) \, dx \, dy + \iint_{A \cap S_r} f(x,y) \, dx \, dy \\ &\le r \iint_A \, dx \, dy + \iint_{S_r} f(x,y) \, dx \, dy \\ &= \iint_{S_r} f(x,y) \, dx \, dy. \end{align} $$ As $r \to \infty$, the set $S_r$ grows smaller and smaller, and so this last integral is also decreasing as $r \to \infty$; it actually decreases to zero because the full integral $\iint_{\mathbb{R}^2} f(x,y) \, dx \, dy$ is finite.