# Existence everywhere of integrand in Fubini's theorem

Let $f\in L(A,\mu_x\otimes\mu_y)$ be a summable function on $A\subset X\times Y$ where $(X\times Y,\mu_x\otimes\mu_y)$ is the product of measure spaces $(X,\mu_x)$ and $(Y,\mu_y)$. Then Fubini's theorem says that $$\int_A f(x,y) d\mu_x\otimes\mu_y=\int_X\Bigg(\int_{A_x}f(x,y)d\mu_y\Bigg) d\mu_x=\int_Y\Bigg(\int_{A_y}f(x,y)d\mu_x\Bigg) d\mu_y.$$

From the proof in Kolmogorov-Fomin's Introductory Real Analysis it seems to me to understand, or misunderstand, that the integrands $\int_{A_x}f(x,y)d\mu_y$ and $\int_{A_y}f(x,y)d\mu_x$ exist everywhere. Nevertheless, further in their original Элементы теории функций и функционального анализа, when talking about the Fredholm operator, it is said (p. 461 here) that $\int_{[a,b]}K(s,t)\varphi(t)d\mu_t$, where $\varphi\in L_2[a,b]$ and $K\in L_2([a,b]^2)$, exists for almost all $s\in[a,b]$.

Do $\int_{A_x}f(x,y)d\mu_y$ and $\int_{A_y}f(x,y)d\mu_x$ respectively exist for all $x\in X$ and $y\in Y$ or do not they? I thank you anybody for any answer!

In general, the inner integral exists only almost everywhere, if $f$ can assume positive, as well as negative values.

To see this, consider e.g.

$$f(x,0) = 1 \text{ if } x >0,\\ f(x,0) = -1 \text{ if } x<0,\\ f(x,y)=0 \text{ otherwise}.$$

Then $f$ is integrable w.r.t. the 2-dimensional Lebesgue measure (on the Borel sigma algebra, which is the product of the 1-dimensional Borel sigma algebras), because it vanishes almost everywhere.

But for $y=0$, the inner integral

$$\int f(x,0)\, dx$$ does not exist.

You do not have this problem for $f \geq 0$, but there the inner integral can be infinite on a null set.

• Now it's clear: the linked proof, without loss of generality, only applies to $f^+$. Thank you so much! – Self-teaching worker Dec 7 '14 at 17:34