Equivlence between integrable function and $J$-measurable set. 
Let $f: A \to \mathbb{R}$ bounded and positive in the block $A \subset \mathbb{R}^{m}$. Then $f$ is integrable if only if the set $C(f) = \{\,(x,y) \in \mathbb{R}^{m+1} \mid x \in A,\; 0 \leq y \leq f(x)\,\}$ be $J$-measurable. In affirmative case,
  $$\int_{A}f(x)\mathrm{d}x = \operatorname{vol}C(f).$$

I don't know how to connect the integrability of $f$ with measure of $C(f)$. I know that $X \subset \mathbb{R}^{m}$ is $J$-measurable if only if $\partial X$ has null measure. So, the question is reduced to 

  
*
  
*$f$ is integrable if only if $\partial C(f)$ is a null set in $\mathbb{R}^{m+1}$.
  

More over, $f$ is integrable if only if $D_{f}$ (discontinuity points of $f$) is a null set. So, we can work with

  
*
  
*$D_{f}$ is null set if only if $\partial C(f)$ is a null set in $\mathbb{R}^{m+1}$.
  

The last version seems to me a little more connected, but still, I don't know how to prove it. Can someone help me?

The function $f$ is integrable if and only if for every $\epsilon > 0$ there is a partition $P$ of $A$ such that $U(P,f) - L(P,f) < \epsilon.$ Notice that the upper/lower sum difference 
$$U(P,f) - L(P,f) = \sum_{j=1}^n (M_j - m_j) \,vol(R_j)$$
is the sum of volumes of a finite collection of rectangles $R_j \times [m_j,M_j]$ that covers the graph $\{(x,f(x)): x \in A\}$ which must, therefore, have measure $0$. The other parts of the boundary of $C(f)$ are sections of $m-$dimensional hyperplanes in $\mathbb{R}^{m+1}$ and obviously have measure $0$.
