Prove that $\int|f − g| = \int_{-\infty}^{\infty} μ(F_t △ G_t) dt.$

Question

Let $f$ and $g$ be integrable functions on a measure space $(X,Σ,μ).$ For each $t ∈ \mathbb{R},$ consider the sets $F_t =\{x∈X :f(x)>t\}, G_t =\{x∈X :g(x)>t\}.$ Prove that $\int|f − g| = \int_{-\infty}^{\infty} μ(F_t △ G_t) dt.$

Here $A △ B = (A $\ B) ∪ (B \ A) denotes symmetric difference of sets.

I am having a lot of difficulty figuring out where to start on this problem. Some direction would be awesome. Its a past qual problem. Thanks.

Answer 1

Notation: For $Y\subset X$, let $1_Y:X\to \Bbb R$ defined by $1_Y(y) = \begin{cases}1 \,\,\,\,\text{ if }y\in Y \\ 0 \,\,\,\,\text{ otherwise.} \end{cases}$

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$$F_t △ G_t = \{ x: f(x) > t \ge g(x)\} \cup \{ x: g(x) > t \ge f(x)\} $$ and the intersection of these sets is empty, so $$\mu(F_t △ G_t) = \mu(\{ x: f(x) > t \ge g(x)\}) + \mu(\{ x: g(x) > t \ge f(x)\}); \\ \int_\Bbb R dt \mu(\{ x: f(x) > t \ge g(x)\}) = \int_X d\mu \left(\int_\Bbb R dt \{ t: f(x) > t \ge g(x)\}\right) \\= \int_X d\mu (f(x)-g(x))1_{f(x)>g(x)} $$ using the Fubini theorem in the middle. Do the same thing with the other part and you are done.