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Spring 2009 - Problem 1

Fubini's theorem

Let $f$ and $g$ be real-valued integrable functions on a measure space $\p{X, \mathcal{B}, \mu}$ , and define

F_t = \set{x \in X \mid f\p{x} > t},\ G_t = \set{x \in X \mid g\p{x} > t}.

Prove

\int \abs{f - g} \,\diff\mu = \int_{-\infty}^\infty \mu\p{\p{F_t \setminus G_t} \cup \p{G_t \setminus F_t}} \,\diff{t}.

Solution.

First, notice that $F_t \setminus G_t$ and $G_t \setminus F_t$ are disjoint, so

\chi_{\p{F_t \setminus G_t} \cup \p{G_t \setminus F_t}} = \chi_{F_t \setminus G_t} + \chi_{G_t \setminus F_t}.

Because $f$ and $g$ are measurable, the sets

A = \set{\abs{f - g} \neq 0} \quad\text{and}\quad A_n = \set{\abs{f - g} \geq \frac{1}{n}}

are measurable. Moreover, because $f$ and $g$ are integrable,

\frac{\mu\p{A_n}}{n} \leq \int_{A_n} \abs{f - g} \,\diff\mu \leq \int_X \abs{f - g} \,\diff\mu < \infty.

Combining this with the fact that $A = \bigcup_{n=1}^\infty A_n$ , we see that $A$ is a $\sigma$ -finite set with respect to $\mu$ . Hence, we may apply Fubini-Tonelli on $\R \times A$ .

Notice that on $x \in A^\comp$ , then $f\p{x} = g\p{x}$ , and so

F_t \cap A^\comp = G_t \cap A^\comp \implies \p{F_t \setminus G_t} \cap A^\comp = \p{G_t \setminus F_t} \cap A^\comp = \emptyset.

In other words, $F_t \setminus G_t \subseteq A$ and $G_t \setminus F_t \subseteq A$ , and so by Fubini-Tonelli,

\begin{aligned} \int_{-\infty}^\infty \mu\p{\p{F_t \setminus G_t} \cup \p{G_t \setminus F_t}} \,\diff{t} &= \int_{-\infty}^\infty \int_X \chi_{F_t \setminus G_t} + \chi_{G_t \setminus F_t} \,\diff\mu \,\diff{t} \\ &= \int_{-\infty}^\infty \int_A \chi_{F_t \setminus G_t} + \chi_{G_t \setminus F_t} \,\diff\mu \,\diff{t} \\ &= \int_A \int_{-\infty}^\infty \chi_{F_t \setminus G_t} + \chi_{G_t \setminus F_t} \,\diff{t} \,\diff\mu \\ &= \int_A \int_{-\infty}^\infty \chi_{\set{g\p{x} \leq t < f\p{x}}} + \chi_{\set{f\p{x} \leq t < g\p{x}}} \,\diff{t} \,\diff\mu \\ &= \int_A \int_{g\p{x}}^{f\p{x}} \chi_{\set{g\p{x} < f\p{x}}} \,\diff{t} + \int_{f\p{x}}^{g\p{x}} \chi_{\set{f\p{x} < g\p{x}}} \,\diff{t} \,\diff\mu \\ &= \int_A \p{f\p{x} - g\p{x}}\chi_{\set{g\p{x} < f\p{x}}} + \p{g\p{x} - f\p{x}} \chi_{\set{f\p{x} < g\p{x}}} \,\diff\mu \\ &= \int_A \abs{f - g} \,\diff\mu \\ &= \int_X \abs{f - g} \,\diff\mu, \end{aligned}

since $\int_{A^\comp} \abs{f - g} \,\diff\mu = 0$ .