$\gdef\uint{\overline{\int}}$ $\gdef\lint{\underline{\int}}$

• Redo Fubini's Theorem (Tao 8.5.1)

To deal with possibly non-integrable functions, we need to define 'upper Lebesgue integral' and 'lower Lebesgue integral', which works for non-integrable functions $\overline{\int} f(x) = \inf \{ \int g(x), \text{g absolutely integrable, and $g(x)>f(x)$} \}$ similarly for lower Lebesgue integral. By monotonicity of integral, we always have upper integral greater than lower integral.

Lemma 8.3.6 says, if a function $f: \R^n \to \R$ satisfies $\uint f = \lint f$, then $f$ is absolutely integrable. To prove it, we create a sequence that approximate $f$ from above, $\overline f_n$ and a sequence that approximate $f$ from below $\underline f_n$, take their limit to get $F_+, F_-$ with $F_+ \geq F_-$. Since $\int F_+ = \uint f = \lint f = \int F_-$, we have $\int F_+ - F_- = 0$, since $F_+ -F_-\geq 0$, we have $F_+ = F_-$ a.e., since $F_+ \geq f \geq F_-$, thus $f = F_+$ a.e., thus measurable and absolutely integrable.

Let $f(x,y): \R^2 \to \R$ be an absolutely integrable function, then there exists integrable function $F(x)$ and $G(y)$, such that for a.e $x$, we have $F(x) = \int f(x,y) dy$ and for a.e $y$, $G(y) = \int f(x,y) dx$, and $\int f(x,y) dx dy = \int F(x) dx = \int G(y) dy$

Pf: We only consider the statement about $F(x)$.

1. We introduce two non-negative functions $f_-, f_+$ (with disjoint support) such that $f = f_+ - f_-$. Then, suppose we prove the theorem for $f_+$ and $f_-$, we can combine the result to get that of $f$. Thus, we only need to deal the case where $f$ is non-negative.
2. Since $f$ is the sup of functions with bounded support, $f = \sup_N f_N, f_N = f \chi_{ [-N, N] \times [-N, N]}$, if $F_N(x) = \int f_N(x,y) dy$ for $x \notin Z_N$, then then we can define $F = \sup_N F_N$, For $x \notin Z = \cup_N Z_N$ (countable union of null-set is still null), by monotone convergence theorem (for upward non-negative functions, we have $F(x) = \sup_N F_N(x) = \sup_N \int f_N(x,y) dy = \int \sup_N f_N(x,y) dy = \int f(x,y) dy$ Hence, suffice to prove the statement for functions with support in a big box $[-N, N] \times [-N, N]$.
3. By since $f$ is $\sup$ of simple functions, we may replace $f$ by simple functions, and go back to $f$ using monotone convergence theorem.
4. Replace simple function by characteristic function, by linearity of integration.
5. Here is the core, we only need to prove the case for $f = 1_E$, where $E \In [-N,N]^2$ is a measurable set. We claim that $\uint (\uint 1_E(x,y) dy) dx \leq m(E)$. The proof of the claim is by box covering, as we did last time (see Tao for detail). Given this claim, we can now finish the proof. Let $E^c = [-N,N]^2 \RM E$, then we have

$4N^2 - \lint (\lint 1_E(x,y) dy )dx = \uint (\uint 1_{E^c}(x,y) dy) dx \leq m(E^c) = 4N^2 - m(E)$ So, $\lint (\lint 1_E(x,y) dy )dx \geq m(E)$ In particular, $\lint (\uint 1_E(x,y) dy )dx \geq \lint (\lint 1_E(x,y) dy )dx \geq m(E) \geq \uint (\uint 1_E(x,y) dy )dx \geq \lint (\uint 1_E(x,y) dy )dx$ Hence $F_+(x) = \uint 1_E(x,y) dy$ is integrable. Similarly $\lint (\lint 1_E(x,y) dy )dx \geq m(E) \geq \uint (\uint 1_E(x,y) dy )dx \geq \uint (\lint 1_E(x,y) dy )dx \geq \lint (\lint 1_E(x,y) dy )dx$ thus $F_- (x) = \lint 1_E(x,y) dy$ is integrable. And, we have $\int F_+(x) dx = \int F_- (x) dx$ hence $F_+(x) = F_-(x)$ for almost all $x$. Thus, for a.e. $x$, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus $\int f(x,y) dy$ exists for a.e. x.