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While reading Folland, I'm running into some trouble understanding the almost everywhere part of Fubini's theorem:

($\bf{Tonelli}$) If $f\in L^+(X,Y)$, then $\displaystyle g:x\mapsto\int_Yf_xd\nu$ is $\mathcal{M}$-measurable,\ $\displaystyle h:y\mapsto \int_Xf^yd\mu$ is $\mathcal{N}$-measurable (so $g\in L^+(X)$ and $h\in L^+(Y)$). And $$\displaystyle \int_{X\times Y}fd\mu \times\nu=\int_Xgd\mu=\int_Yhd\nu.$$ That is, $$\displaystyle \int_{X\times Y}fd\mu \times\nu=\int_X\left(\int_Yf_xd\nu\right)d\mu(x)=\int_Y\left(\int_Xf^yd\mu\right)d\nu(y)$$ ($\bf{Fubini}$) If $f\in L^1(X\times Y)$, then $f_x\in L^1(Y,\nu)$ for a.e. $x\in X$ and $f^y\in L^1(X,\mu)$ for a.e. $y\in Y$. The a.e. defined functions $g$ and $h$ above are $\mathcal{M}$-measurable and $\mathcal{N}$-measurable respectively and the conclusion from above holds.

For now, assume Tonelli's part is true (which I don't have a huge problem proving). Then Since $f\in L^1(X\times Y)$, then $|f|\in L^+$. That is $(\int_{X\times Y})|f|<\infty$. Then

$\Rightarrow$ $\displaystyle\int_{X\times Y}|f|d\mu \times\nu=\int_Y\left(\int_X|f^y|d\mu\right)d\nu=\int_X\left(\int_Y|f_x|d\nu\right)d\mu$

$\Rightarrow \displaystyle \int_X|f^y|d\mu<\infty$ a.e. $y$ and $\displaystyle \int_Y|f_x|d\nu<\infty$ a.e. $x$.

and this is where I am stuck. I'm not sure about this last line and how it follows from the above work. Any help would be great.

Answer: In Folland use Theorem 2.20 for $|f^y|$ and $|f_x|$

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If $\int_X g(x) d\mu(x) < \infty$, then the integrand $g(x)$ is finite $\mu$-almost everywhere. In your case, the integrand is taken to be $\int_Y |f_x| d\nu$, and so this must be finite for $\mu$-a.e. $x$. The analogous statement holds for $\int_X |f^y | d\mu$ after changing the order in which the product is decomposed.

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  • $\begingroup$ I had forgotten about this theorem. Thank you. $\endgroup$ – user23793 Jan 6 '15 at 5:50

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