Why can we use MCT in the proof of Fubini's theorem (in Stein's book)?

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In Stein's Real Analysis, he uses six steps to prove Fubini's theorem.

First he uses $\mathcal F$ to denote the set of integrable funcyions satisfying the conclusions in the theorem:

$f\in L^1(\mathbb R^{d_1}\times \mathbb R^{d_2})$ and

  1. $f^y\in L^1(\mathbb R^{d_1})$ a.e. $y\in \mathbb R^{d_2}$.

  2. $\int_{\mathbb R^{d_1}}f^y(x)\mathrm d x\in L^1(\mathbb R^{d_2})$.

  3. $\int_{\mathbb R^{d_2}}\int_{\mathbb R^{d_1}}f(x,y)\mathrm dx\mathrm dy=\int_{\mathbb R^d}f.$

However, it confuses me that the definition of $\mathcal F$ is unclear, because it's not trivial that if $f\in\mathcal F$ and $f=g$ a.e., then we have $g\in \mathcal F$.

Besides, when proving that $\mathcal F$ is closed under operation of limits, he assumes that $\mathcal F\ni f_k\nearrow f$(the $\nearrow$ sign implies "$f_k\to f$ increasingly a.e." in the book) Then he applies MCT on $f_k^y$, but I can't see why $f_k^y\nearrow f^y$ is true? (If we have $\lim f_k=f$ for every point in $\mathbb R^d$, then we can concludes that $f^y_k$ is integrable except a set $A(m(A)=0)$, then for all $y\in A^c$, $f^y_k\nearrow f^y$ is true. But here we must prove when $f_k\nearrow f$ , we also have $f^y_k\nearrow f^y$?)

I think the two questions above can both be solved if we can prove that any function supported on a set of measure zero is in $\mathcal F$, so we can drop the condition "a.e." to prove $f^y_k\to f^y$ increaingly everwhere. Hence it's sufficient to prove that if $m(E)=0$, then $m(E^y)=0$ a.e. $y\in \mathbb R^{d_2}$? (But Stein proves this fact on step 4, which relies on the result we just want to prove.)

I don't know if I have missed something important or misunderstood the author, please point out any mistakes or give any hints.