# Fubini theorem makes no sense then points are not measurable

I was reviewing Fubini theorem for general measures and noticed that I need $\{x\} \in \mathcal{F}_{X}$ to complete the proof:

Do I need $\{ x\} \in \mathcal{F}_X$ for all $x$ in order to state Fubini-(Toneli) theorem?

General Fubini theorem states that given $\sigma$-finite measure space $(X,\mathcal{F}_X,\mu)$, measurable space $(Y,\mathcal{F}_Y)$, a family of measures $\nu_x$ over $(Y,\mathcal{F}_Y)$ parametrized by elements of $X$ satisfying certain conditions, and measurable function $f$ over $(X \times Y,\mathcal{F}_X \times \mathcal{F}_Y)$ which is nonnegative (or has integral or integrable) the function of $x \in X$ $$\int_Y f(x,y) \, d\nu_x(y)$$ is measurable, and

$$\int_X \int_Y f(x,y) \, d \nu_x(y) \, d\mu(x) = \int_{X \times Y}f \, d \nu'$$ where $\nu'$ is uniquely defined by $\nu_x$ and $\mu$

However, to make this theorem make sense we need to show for all $x$ that $f(x,\cdot)$ is measurable with respect to $\mathcal{F}_Y$. At first this seems trivial. Take Borel set $A \subset \mathbb{R}$. Then

$$(f(x, \cdot))^{-1}(A) = \pi_Y(f^{-1}(A) \cap \{x\} \times Y)$$

where $\pi_Y$ is a projection $\pi_Y : (a,b) \mapsto b$.

$f^{-1}(A)$ is measurable set by definition of $f$ and if $\{x\} \in \mathcal{F}_X$ the whole intersection will be measurable which is the case then $X = \mathbb{R}^n$ wiht Borel $\sigma$-algebra.

However intersection may happen to be non-measurable then $\{x\} \not \in \mathcal{F}_X$. So what to do in this case?

Call measurable space $(X, \mathcal{F}_X)$ such that $\{x\} \in \mathcal{F}_X$ for all $x \in X$ fubinible . Do I need to ask for $(X,\mathcal{F}_X)$ to be fubinible is statement of Fubini theorem? Are there other version of Fubini theorem which don't requere fubinible measurable space, say for point-free lattices?

The product $\sigma$-algebra $\mathcal{F}_X \times \mathcal{F}_Y$ has the property that all sections of measurable sets belong to the corresponding factor $\sigma$-algebra. That is, we have

$$A_x := \{ y \in Y : (x,y) \in A\} \in \mathcal{F}_Y\tag{\ast}$$

for all $A \in \mathcal{F}_X \times \mathcal{F}_Y$ and all $x\in X$, and the analogous statement with the roles of $X$ and $Y$ exchanged.

From $(\ast)$ it easily follows that each section $f(x,\,\cdot\,)$ is $\mathcal{F}_Y$-measurable for $\mathcal{F}_X\times \mathcal{F}_Y$-measurable $f$, since $\bigl(f(x,\,\cdot\,)\bigr)^{-1}(A) = \bigl(f^{-1}(A)\bigr)_x \in \mathcal{F}_Y$ for Borel sets $A$ by $(\ast)$.

Thus let's prove $(\ast)$. For every $x \in X$, define

$$\mathcal{S}_x := \{ A \subset X\times Y : A_x \in \mathcal{F}_Y\}.$$

Then since $\varnothing_x = \varnothing$, $(X\times Y\setminus A)_x = Y \setminus A_x$ and $\bigl(\bigcup A_n\bigr)_x = \bigcup (A_n)_x$, it follows that $\mathcal{S}_x$ is a $\sigma$-algebra. Also, we have $U\times V \in \mathcal{S}_x$ for all $U \in \mathcal{F}_X$ and $V\in \mathcal{F}_Y$, since $(U\times V)_x$ is either $\varnothing$ or $V$. By definition,

$$\mathcal{F}_X \times \mathcal{F}_Y = \sigma\bigl(\{ U \times V : U \in \mathcal{F}_X,\, V \in \mathcal{F}_Y\}\bigr)$$

is the smallest $\sigma$-algebra containing all measurable rectangles, so

$$\mathcal{F}_X \times \mathcal{F}_Y \subset \mathcal{S}_x.$$

But $x\in X$ was completely arbitrary, so

$$\mathcal{F}_X \times \mathcal{F}_Y \subset \bigcap_{x\in X} \mathcal{S}_x,$$

which is just $(\ast)$.

• Thank you Daniel Fischer! At first I thought that proof will require axiom of choice but now as I see that inverse is just a section everything is clear. – Nik Pronko Nov 23 '16 at 16:48