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Spring 2014 - Problem 1

Fubini-Tonelli, measure theory

Consider a measure space $\p{X, \mathcal{X}}$ with a sigma-finite measure $\mu$ and, for each $t \in \R$ , let $e_t$ denote the characteristic function of the interval $\p{t, \infty}$ . Prove that if $\func{f,g}{X}{\R}$ are $\mathcal{X}$ -measurable, then $\norm{f - g}_{L^1\p{X}} = \int_\R \norm{e_t \circ f - e_t \circ g}_{L^1\p{X}} \,\diff{t}.$

Solution.

First, notice that

\abs{e_t \circ f - e_t \circ g}\p{x} = \begin{cases} 1 & \text{if } f\p{x} \leq t < g\p{x}, \\ 1 & \text{if } g\p{x} \leq t < f\p{x}, \\ 0 & \text{otherwise}. \end{cases}

We begin on the right-hand side. Everything is non-negative and all measures are $\sigma$ -finite, so we may apply Fubini-Tonelli to get

\begin{aligned} \int_\R \norm{e_t \circ f - e_t \circ g}_{L^1} \,\diff{t} &= \int_\R \int_X \abs{e_t \circ f - e_t \circ g}\p{x} \,\diff\mu\p{x} \,\diff{t} \\ &= \int_\R \int_{\set{f < g}} \chi_{\set{f\p{x} \leq t < g\p{x}}} \,\diff\mu\p{x} + \int_{\set{g < f}} \chi_{\set{g\p{x} \leq t < f\p{x}}} \,\diff\mu\p{x} \,\diff{t} \\ &= \int_{\set{f < g}} \int_\R \chi_{\set{f\p{x} \leq t < g\p{x}}} \,\diff{t} \,\diff\mu\p{x} + \int_{\set{g < f}} \int_\R \chi_{\set{g\p{x} \leq t < f\p{x}}} \,\diff{t} \,\diff\mu\p{x} \\ &= \int_{\set{f < g}} \int_{f\p{x}}^{g\p{x}} \,\diff{t} \,\diff\mu\p{x} + \int_{\set{g < f}} \int_{g\p{x}}^{f\p{x}} \,\diff{t} \,\diff\mu\p{x} \\ &= \int_{f < g} \abs{f\p{x} - g\p{x}} \,\diff\mu\p{x} + \int_{g < f} \abs{f\p{x} - g\p{x}} \,\diff\mu\p{x} \\ &= \int_X \abs{f\p{x} - g\p{x}} \,\diff\mu\p{x} \\ &= \norm{f - g}_{L^1}. \end{aligned}