# How can I use Fubini's Theorem here?

Exercise:

If $$F$$ is a continuous distribution function on $$(\mathbb R, \mathscr B, \mu_{\mathcal L})$$ with distribution $$\mu_F$$, use Fubini's theorem to show that

1. $$\int_{\mathbb R} F(x) \, d\mu_F(x) = \frac{1}{2}$$

2. if $$X_1, X_2$$ are i.i.d random variables with common distribution $$F$$, then $$P(\{X_1 \leq X_2 \}) = 1/2$$ and $$\text E(F(X_1)) = 1/2$$.

My Attempt:

I don't really understand bs_math's answer so I have been trying to write my own. I just deleted an attempt here that was (I think) completely nonsensical. I am working on another attempt.

For example, I don't understand what's going on in line 4 of bs_math's answer.

• Hint: Write $F(x)$ in terms of $\mu_F$ Commented Mar 25, 2022 at 18:19
• @bs_math I have a case-by-case definition stating that \begin{align*} F(x) = \begin{cases} \mu((0, x]) \text{ if } x > 0\\ 0 \text{ if } x = 0\\ -\mu((x, 0]) \text{ if } x < 0. \end{cases} \end{align*} You think this leads to Fubini's theorem? Commented Mar 25, 2022 at 18:28
• Unless I'm mistaken that would give me two integrals that look something like $\int \mu_F\big((0, x] \big) \, d\mu_f$, and I am not sure what the integral of a measure with respect to that same measure is. Commented Mar 25, 2022 at 18:41
• Try $\mu((0, x]) = \int_{-\infty}^x 1 d\mu_F(x)$ Commented Mar 25, 2022 at 18:44
• @bs_math Thanks. I need to try to remember how the measure relates to an improper Riemann integral now. Commented Mar 25, 2022 at 18:51

The distribution function is generally defined as $$F: x \mapsto \mu_F((-\infty, x]),$$ see Wikpedia. The OP relies on another definition stating that $$\tilde{F}(x) = \begin{cases} \mu_F((0, x], & x > 0, \\ 0, &x = 0, \\ - \mu_F((-x, 0], &x < 0. \end{cases}$$ Then we have $$\tilde{F} = F + c$$ for the constant summand $$c = - \mu_F((-\infty, 0])$$. I have never seen this as a definition of the distribution function. The fact that one can show the claim $$\int_\mathbb{R} F(x) \, d\mu_F(x) = \frac{1}{2} \tag{1}$$ for $$F$$ shows that $$(1)$$ is actually false as soon as $$c \neq 0$$. So I assume there is a misunderstanding regarding the intended definition of $$F$$.
Now, in order to show $$(1)$$ use that by the very definition of the Lebesgue integral it is $$F(x) = \mu((-\infty, x]) = \int_\mathbb{R} 1_{(-\infty, x]}(y) \, d\mu_F(y).$$ Then, consider the following transformations: \begin{align} \int_\mathbb{R} F(x) d\mu_F(x) &= \int_{\mathbb{R}} \int_\mathbb{R} 1_{(-\infty, x]}(y) \, d\mu_F(y) \, d\mu_F(x) \\ &= \int_\mathbb{R} \int_\mathbb{R} 1_{(-\infty, x]}(y) \, d\mu_F(x) \, d\mu_F(y) \\ &= \int_\mathbb{R} \int_{\mathbb{R}} 1_{[y, \infty)}(x) \, d \mu_F(x) \, d\mu_F(y) \\ &= \int_\mathbb{R} 1 - F(y) \, d\mu_F(y) \\ &= 1 - \int_\mathbb{R} F(x) \, d \mu_F(x), \tag{2} \end{align} where we use Fubinis's theorem in the second line. To get from the second to the third line we observe for every $$x, y \in \mathbb{R}$$ $$1_{(-\infty, x]}(y) = 1 \Leftrightarrow y \leq x \Leftrightarrow 1_{[y, \infty)}(x) = 1.$$ In the fourth line we use that for fixed $$y$$ one has \begin{align*} \int_\mathbb{R} 1_{[y, \infty)}(x) \, d \mu_F(x) &= \int_\mathbb{R} 1 - 1_{(-\infty, y)}(x) \, d\mu_F(x) \\ &= 1 - \int_\mathbb{R} 1_{(-\infty, y)}(x) \, d\mu_F(x) \\ &= 1 - \lim_{h \searrow 0} \int_\mathbb{R} 1_{(-\infty, y - h]}(x) \, d\mu_F(x) \\ &= 1 - \lim_{h \searrow 0} F(y-h) \\ &= 1 - F(y), \end{align*} where we use Beppo-Levi's theorem in the third line.
The claim $$(1)$$ follows immediately from $$(2)$$ by rearranging terms.
• If you have time I'd appreciate an explanation for what you've written. I don't understand your first line. What about the three cases for $F$? Can the inner integral on the first line be written as a Lebesgue integral? Why is it okay to rewrite the indicator function on line 4? How does line 5 follow from line 4 without a density in the integrand? Thanks. Commented Mar 26, 2022 at 4:51