Q:
Do we have a proof that Fubini's Theorem is True?
I was wondering if the proof of Fubini's Theorem is a proof that uses Fubini's Theorem? If so, then isn't it circular?
A:
Let $A \subset \mathbb{R}^{n}$, $B \subset \mathbb{R}^{m}$ be measurable sets, then
$x\in A,y\in B \Rightarrow A\times B\cap(A\times B+ (x,y)) $ is measurable
If we define $f(x,y) =1$ if $(x,y)\in A\times B$ and $0$ otherwise, we have the following:
$$\int_{A\times B} f(x,y) d x dy = \int_{A\times B} f(x,y) d y dx = \int_{A} \left( \int_{B} f(x,y) dy \right) d x= \int_{A} \left( \int_{B} f(x,y) d x\right) dy = \int_{A} \left( \int_{B} f(x,y) dx\right) dy$$
The penultimate equality is the Fubini Theorem.
A:
Fubini's theorem is used in the proof of the theorem itself.
A:
No, we use Fubini's theorem, but we do not need to use it to prove that it is true. We can prove it is true directly using simple algebraic methods.
Just assume Fubini's theorem and prove it holds, then deduce Fubini's theorem from the result we have proved.
We would be proving it by starting with the following statement.
If $A$ and $B$ are measurable sets, then the set of all ordered pairs $(x,y)$ of real numbers such that $x\in A$ and $y\in B$ has Lebesgue measure zero.
First, if $(x,y)$ belongs to the set, then $x$ must belong to $A$, and therefore $x$ must belong to the union of all sets of the form $A+m$, where $m$ belongs to $\mathbb Z$, because every real number is equivalent to a integer in this regard. Similarly, $y$ must belong to the union of all sets of the form $B+n$, where $n$ belongs to $\mathbb Z$. Therefore, $(x,y)$ must belong to the union of all sets of the form
$$
A+B+n+m = \{ a + b + n + m: a\in A, b\in B, n,m\in\mathbb Z \}.
$$
The following lemma holds:
$E \subset \mathbb R^2$ is of the form $A+B+n+m$, where $A$ and $B$ are subsets of $\mathbb R$, if and only if $E$ is a subset of the union of the sets $\{ (i,j): i\in A, j\in B \} \cup \{ (0,k): k\in \mathbb Z \} \cup \{ (-k,0): k\in \mathbb Z \} \cup \{ (0,-k): k\in \mathbb Z \} \cup \{ (-i,j): i\in \mathbb R, j\in B \} \cup \{ (i,0): i\in A \}$.
The union of the sets $\{ (i,j): i\in A, j\in B \} \cup \{ (0,k): k\in \mathbb Z \} \cup \{ (-k,0): k\in \mathbb Z \} \cup \{ (0,-k): k\in \mathbb Z \} \cup \{ (-i,j): i\in \mathbb R, j\in B \} \cup \{ (i,0): i\in A \}$ is equal to $\mathbb R^2$, because if $k$ and $l$ are natural numbers, then so is $k+l$. (If $k$ and $l$ are not natural numbers, then it is clear that $(k,l)$ is not in the union.)
Since the union of two sets is a subset of the union of a larger set, we have proven:
A set $E$ of the form $A+B+n+m$ is a subset of the union of the sets $\{ (i,j): i\in A, j\in B \} \cup \{ (0,k): k\in \mathbb Z \} \cup \{ (-k,0): k\in \mathbb Z \} \cup \{ (0,-k): k\in \mathbb Z \} \cup \{ (-i,j): i\in \mathbb R, j\in B \} \cup \{ (i,0): i\in A \}$.
By Fubini's theorem, the left side of this statement is equal to
$$
\int_{\mathbb R^2} 1_E(x,y) dx dy.
$$
Now, if $(x,y)\notin E$, then $x$ is not in $A$, $y$ is not in $B$, $x\notin A + n$, and $y\notin B+n$, or $x\in A+n$ and $y\notin B+n$, for any natural number $n$. Therefore, $x$ is not in the union of all sets of the form $A+k$, where $k$ belongs to $\mathbb Z$, or $y$ is not in the union of all sets of the form $B+k$, where $k$ belongs to $\mathbb Z$, and hence $(x,y)\notin E + \mathbb Z^2$, where $\mathbb Z^2$ is the union of all sets of the form
$$
\left\{ (x,y): x\in A+k, y\in B+l, k,l\in\mathbb Z \right\}.
$$
In other words, $(x,y)\notin E + \mathbb Z^2$ if and only if one of the following holds:
$(x,y)\notin E$;
$x\notin A$;
$y\notin B$;
$x\in A$;
$y\in B$;
$x\in A+k$ and $y\notin B+l$, where $k,l$ belong to $\mathbb Z$;
$x\in B+m$ and $y\in A+n$, where $m,n$ belong to $\mathbb Z$.
Using this, we see that the integral of $1_E$ over $\mathbb R^2$ is equal to zero. Therefore, the set of all ordered pairs $(x,y)$ such that $x\in A$ and $y\in B$ has Lebesgue measure zero. Since $(x,y)$ can be any ordered pair of real numbers, $A$ and $B$ have measure zero.
Using the above notation, we have proven:
$$
\int_{\mathbb R^2} 1_E(x,y) dx dy = \int_{A+B+n+m} 1_E(x,y) dx dy = \int_{A+B} 1_E(x,y) dx dy = \int_{A} \left( \int_{B} 1_E(x,y) dy \right) dx.
$$
The result is true, if and only if, for any Lebesgue measurable set $A$ and $B$, the double integral
$$
\int_{A} \left( \int_{B} 1_E(x,y) dy \right) dx
$$
is equal to zero, or in other words,
$$
\int_{A+B} 1_E(x,y) dx dy = 0
$$
whenever $(x,y)$ belongs to the union of all sets of the form
$$
\{ (i,j): i\in A, j\in B \} \cup \{ (0,k): k\in \mathbb Z \} \cup \{ (-k,0): k\in \mathbb Z \} \cup \{ (0,-k): k\in \mathbb Z \} \cup \{ (-i,j): i\in \mathbb R, j\in B \} \cup \{ (i,0): i\in A \}.
$$
