The integral of a characteristic function with respect to a product measure.

Problem: Let $ (X,\mathcal{A},\mu) $ and $ (Y,\mathcal{B},\nu) $ be measure spaces, where

• $ X = Y $ is the interval $ [0,1] $,
• $ \mathcal{A} = \mathcal{B} $ is the collection of Borel subsets of $ [0,1] $,
• $ \mu $ is the Lebesgue measure and $ \nu $ is the counting measure, both on $ [0,1] $.

Show that the diagonal set $ \Delta \stackrel{\text{def}}{=} \{ (x,y) \in X \times Y \mid x = y \} $ is measurable with respect to the product measure $ \mu \otimes \nu $. (In fact, it is an $ F_{\sigma \delta} $-set.)

Show also that if $ f $ is the characteristic function of $ \Delta $, then $$ \int_{X \times Y} f ~ \mathrm{d}{(\mu \otimes \nu)} \neq \int_{X} \left[ \int_{Y} f(x,y) ~ \mathrm{d}{\nu(y)} \right] \mathrm{d}{\mu(x)}. $$

Is this a contradiction of either Fubini’s Theorem or Tonelli’s Theorem?

Solution: For each $ n \in \mathbb{N} $ and each $ j \in \{ 1,\ldots,n \} $, let $ I_{j,n} $ denote the interval $ \left[ \dfrac{j - 1}{n},\dfrac{j}{n} \right] $. Also, for each $ n \in \mathbb{N} $, let $ I_{n} $ denote the union $ \displaystyle \bigcup_{j = 1}^{n} I_{j,n} $. Then we have $ \displaystyle \Delta = \bigcap_{n = 1}^{\infty} I_{n} $, so $ \Delta $ is measurable.

Next, observe that $$ \int_{X} \left[ \int_{Y} {\chi_{\Delta}}(x,y) ~ \mathrm{d}{\nu(y)} \right] \mathrm{d}{\mu(x)} = \int_{X} \nu(\{ y \in Y \mid y = x \}) ~ \mathrm{d}{\mu(x)} = \int_{X} 1 ~ \mathrm{d}{\mu} = 1, $$ whereas $$ \int_{Y} \left[ \int_{X} {\chi_{\Delta}}(x,y) ~ \mathrm{d}{\mu(x)} \right] \mathrm{d}{\nu(y)} = \int_{Y} \mu(\{ x \in X \mid x = y \}) ~ \mathrm{d}{\nu(y)} = \int_{Y} 0 ~ \mathrm{d}{\nu} = 0. $$

Now, suppose that $ (A_{n})_{n \in \mathbb{N}} $ and $ (B_{n})_{n \in \mathbb{N}} $ are sequences of Borel subsets of $ [0,1] $ such that $$ \Delta \subseteq \bigcup_{n = 1}^{\infty} (A_{n} \times B_{n}). $$ We can find an $ N \in \mathbb{N} $ such that $ B_{N} $ is infinite, which gives us $ (\mu \otimes \nu)(A_{N} \times B_{N}) = \infty $. Hence, $$ \sum_{n = 1}^{\infty} (\mu \otimes \nu)(A_{n} \times B_{n}) = \infty. $$ The definition of an outer measure therefore yields $$ \int_{X \times Y} \chi_{\Delta} ~ \mathrm{d}{(\mu \otimes \nu)} = (\mu \otimes \nu)(\Delta) = \infty. $$

Solution 1:

There is no contradiction with Fubini's or Tonelli theorem because one of the measure is not $\sigma$-finite (namely, counting measure on the unit interval, an uncountable set). Note that the assumption of non-negativeness is satisfied.

It's a good example to remember in order to not forget any assumption in these theorems.