Theorem: (I am not sure how this theorem is called but it is similar to Fubini's theorem) If $f(x,y)$ is non-negative and measurable(in the Lebesgue way) on $\Bbb R^{n+m}$.Then for every $x_0$ (for a fixed $x$) belonging to $\Bbb R^n$, the function $f(x_0,y)$ is measurable for $y$ belonging to $\Bbb R^m$: $I(x)=∫f(x,y)\,dy$ (on $\Bbb R^m$) is a measurable function of $x$ and $∫f(x,y)\,dx\,dy=∫I(x)\,dx$ (on $\Bbb R^{n+m}=\Bbb R^n$) is true.
My question is: if $g(y)$ is a non-negative integrable function on$\Bbb R^m$ so that $∫g(y)\,dy=0$,why is $f(x,y)=g(y)$ not integrable on $\Bbb R^{n+m}$ although the double integral exists?
Though I don't think that I understand the question correctly, the theorem by Fubini and Tonelli says that if $f(x,y)$ is a non-negative function measurable with respect to the product space $X\times Y$ (with the product sigma-algebra and the product measure), then you can always interchange the order of integration. That is, $\int_{X\times Y}f(x,y)d(\mu \times \nu)=\int_{X}(\int_{Y}f(x,y)d\nu)d\mu=\int_{Y}(\int_{X}f(x,y)d\mu)d\nu$, where I have taken the measure on $X$ to be $\mu$ and on $Y$ to be $\nu$. Also, the above equality holds in the extended sense, so if any one of them is finite then $f$ is integrable. In your question, if we use this theorem for $f$, then each vertical slice of $f$, that is, $g$ integrates to $0$. Hence, $\int_{\mathbb{R}^n\times \mathbb{R}^m} f=\int_{\mathbb{R}^n}0=0$. Hence, $f$ is integrable on $\mathbb{R}^{n+m}$ as $0<\infty$. Actually, $f$ is $0$ almost everywhere.