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Let $A\subset \mathbb R^m$ be a rectangle and $f:A\to \mathbb R$ be a bounded integrable function. I'm trying to prove the graph of $f$ has volume zero.

We define the volume of a J-measurable set $X$ as $\int_A\chi_X(x)$ where $A$ is a rectangle and $\chi_X$ is the characteristic function, i.e., $\chi_X(x)=1$ if $x\in X$ and $\chi_X(x)=0$ if $x\in A-X$.

I have to prove this characteristic function is integrable and its integral is zero. I don't have any idea how to proceed. I need a hint to help me to work through this problem

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    The graph is a subset of $\Bbb{R}^{m+1}$ so you should be looking at its $(m+1)$-dimensional volume. If your definition of integrability involves upper and lower sums approaching each other, then the easy hint is to observe that the graph is squeezed in-between. – 2017-01-23
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    @JyrkiLahtonen My problem is I don't know how the volume of the graph looks like. – 2017-01-23
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    Ok. Let's start with the easy case that $f$ is a constant, i.e. $f(x)=a$ for all $x\in A$. Then the graph is a subset of the $(m+1)$-dimensional rectangle $A\times [a-\epsilon,a+\epsilon]$ for all $\epsilon>0$. So the volume of the graph is smaller than the volume of that rectangle. But that rectangle has volume $2\epsilon\ \times$ the volume of $A$. This can be made as small as we wish, so the volume of that graph is zero. In general you need to approximate $f$ by finitely many pieces such that $f$ is constant on all the pieces (or the piece is small itself). – 2017-01-23
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    [Some useful comments](http://math.stackexchange.com/q/220390/11619). – 2017-01-23
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    @JyrkiLahtonen thank you, I got your hint. I'm working on the general case. – 2017-01-23

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