I'm reading the book on measure theory by Axler, https://measure.axler.net/MIRA.pdf.
I'm trying to understand the proof of Fubini's theorem in p.$132$.
Fubini's theorem
Let $(X,\mathscr S,\mu),(Y,\mathcal T,\nu)$ be $\sigma$-finite measure spaces, and $f:X\times Y\to[-\infty,\infty]$ be $\mathscr S\otimes\mathcal T-$ measurable, and suppose $\int_{X\times Y}|f|(\mu\times\nu)<\infty.$
Then, we get
$(1)$ $\int_Y|f(x,y)|d\nu(y)<\infty$ for a.e.$x\in X$
$(2)$ The function $x\mapsto \int_Yf(x,y)d\nu(y)$ is $\mathscr S-$ measurable on $X.$
The similar statements about $y\in Y$ also hold, and
$$(3) \int_{X\times Y}fd(\mu\times\nu)=\int_X\int_Yf(x,y)d\nu(y)d\mu(x)=\int_Y\int_Xf(x,y)d\mu(x)d\nu(y)$$
holds.
I'm reading the proof but I wonder this is not sufficient.
The outline of the proof is here.
Since the function $x\mapsto \int_Y|f(x,y)|d\nu(y)$ is $\mathscr S$-measurable(This is already proved.), I have $$\{x\in X\mid \int_Y|f(x,y)|d\nu=\infty\}\in\mathscr S,$$ so we can consider $\mu(\{x\in X\mid \int_Y|f(x,y)|d\nu=\infty\}).$
From the tonelli's theorem,
$$\int_X\int_Y|f(x,y)|d\nu(y)d\mu(x)=\int_{X\times Y}|f|d(\mu\times\nu)<\infty,$$ and then $\mu(\{x\in X\mid \int_Y|f(x,y)|d\nu(y)=\infty\})=0.$
Note we can write $f=f^+-f^-.$
From $f^+\leqq|f|$ and $f^-\leqq|f|$, we can see that $\{x\in X\mid \int_Yf^+(x,y)d\nu(y)=\infty\}$ and $\{x\in X\mid \int_Yf^-(x,y)d\nu(y)=\infty\}$ have $\mu-$ measure zero.
Thus, the intersection of these two sets, which is the set of $x\in X$ such that $\int_Yf(x,y)d\nu(y)$ is not defined, also has $\mu$– measure zero. In other words, $\mu(\{x\in X\mid \int_Yf(x,y)d\nu(y) \mathrm{is \ not \ defined ,}\})=\mu(\{x\in X\mid \int_Yf^+=\infty \mathrm{and} \int_Yf^-=\infty\})=0.$
Let $\mathcal F:X\to[-\infty,\infty]$ by
$$\mathcal F(x)=
\begin{cases}
\int_Y f(x,y)d\nu(y) &\mathrm{if} \ \int_Y f(x,y)d\nu(y)\ \mathrm{is \ defined}\\
0 &\mathrm{otherwise}
\end{cases}$$
Since two functions $x\mapsto \int_Yf^+(x,y)d\nu(y)$ and $x\mapsto \int_Yf^-(x,y)d\nu(y)$ are $\mathscr S$-measureble(This is already proved.), $\mathcal F$ is also $\mathscr S-$ measurable. (Note that $\int_Yf=\int_Yf^+-\int_Yf^-$.)
Finally, recalling the definition of Lebesgue integral for general functions and applying Tonelli's theorem to $f^+$ and $f^-$, and noting linearity of integral, we get $$\int_{X\times Y}fd(\mu\times\nu)=\int_X\int_Yf(x,y)d\nu(y)d\mu(x).$$ The similar proceduce enable us to prove the statement for $y\in Y.$ Q.E.D.
I wonder this proof doesn't show the statement $(2)$.
The statement $(2)$ in the theorem is "The function $x\mapsto \int_Yf(x,y)d\nu(y)$ is $\mathscr S-$ measurable," but in this proof, what we show is
"The function $x\mapsto\begin{cases}
\int_Y f(x,y)d\nu(y) &\mathrm{if} \ \int_Y f(x,y)d\nu(y)\ \mathrm{is \ defined}\\
0 &\mathrm{otherwise}
\end{cases}$ is $\mathscr S-$ measurable."
In the first place, depending on $x\in X$, $\int_Yf(x,y)d\nu(y)$ can be undefined, so I wonder that the function $x\mapsto \int_Yf(x,y)d\nu(y)$ is not well-defined and the statement $(2)$ in the theorem "The function $x\mapsto \int_Yf(x,y)d\nu(y)$ is $\mathscr S-$ measurable" cannot hold.
Perhaps, I misunderstand the proof somewhere ?
Could you explain about this ?
Best Answer
You're right, the statement of the theorem is a little inexact. What they meant by (2) is "there exists an $\mathcal S$-measurable function $g(x),$ which is almost-everywhere equal to $\int_Y f(x,y)dv(y)$". You can check that this follows from the result you quoted, together with (1).