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

• Redo Fubini's Theorem (Tao 8.5.1)
• Vitali Covering Lemma.

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.

Suppose $A$ is measurable, and $B \In A$ any subset, with $B^c = A \RM B$. Then $$ m(A) = m^*(B) + m_*(B^c) $$ Proof: $$ m^*(B) = \inf \{ m(C) \mid A \supset C \supset B, C \text{measurable} \} = \inf \{ m(A) - m(C^c) \mid A \supset C \supset B, C \text{measurable} \} $$ $$ = m(A) - \sup \{ m(C^c) \mid A \supset C \supset B, C \text{measurable} \} = m(A) - \sup \{ m(C^c) \mid C^c \subset B^c, C^c \text{measurable} \} = m(A) - m_*(B^c) $$

$\gdef\vcal{\mathcal{V}}$ A Vitali covering $\vcal$ of a set $A \In \R^n$ is such that, for any $p \in A, r > 0$, there is a covering set $V \in \vcal$, such that $\{p\} \subsetneq V \In B_r(p)$, where $B_r(p)$ is the open ball of radius $r$ around $p$.

Vitali Covering Lemma: Let $\vcal$ be a Vitali covering of a measurable bounded subset $A$ by closed balls, then for any $\epsilon>0$, there is a countable disjoint subcollection $\vcal' = \{V_1, V_2, \cdots \}$, such that $A \RM \cup_k V_k$ is a null set, and $\sum_k m(V_k) \leq m(A) + \epsilon$.

Proof: The construction is easy, like a 'greedy algorithm'. First, using the given $\epsilon$, we find an open subset $W \supset A$, with $m(W) \leq m(A) + \epsilon$. Let $\vcal_1 = \{V \in \vcal: V \In W\}$, and $d_1 = \sup \{diam V: V \in \vcal_1\}$. We pick $V_1 \in \vcal_1$ where the diameter is sufficiently large, say $diam V_1 > d_1 /2$. Then, we delete $V_1$ from $W$, let $W_2 = W \RM V_1$, and consider $\vcal_2 = \{ V \in \vcal_1, V \In W_2\}$, and define $d_2 = \sup \{diam V: V \in \vcal_2\}$, and pick $V_2$ among $\vcal_2$ so that $diam V_2 > d_2 /2$. Repeat this process, we get a collection of disjoint closed balls $\{V_i\}$. Suffice to show that $A \RM \cup V_i$ is a null set.

The crucial claim is the following, for any positive integer $N$, we have $$ \cup_{k=N}^\infty 5 V_k \supset A \RM (\cup_{i=1}^{N-1} V_i) $$ Suppose not, and there is a point $a \in A \RM (\cup_{i=1}^{N-1} V_i)$, but not in $\cup_{k=N}^\infty 5 V_k$, then we can find a closed ball $B \in \vcal_N$, such that $a \in B$. Since $a \notin 5V_N$, we have $B \not\subset 5V_N$. This implies $B \cap V_N = \emptyset$. Draw a picture. This implies $B \in \vcal_{N+1}$. Then, repeat the above story $N$ replaced by $N+1$ and same $a,B$, we can keep going and show that $B \in \vcal_k$ for all $k \geq N$. That cannot be true, since $d_k \to 0$, but $B$ has fixed radius.