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Question: Let $f$ and $g$ be Lebesgue integrable function on $\mathbb{R}^d$. Show that $$\int_{\mathbb{R}^d} (f(x) - g(x))^+ \, dx = \int_{-\infty}^{\infty} |\{g(x) < t < f(x)\}|\, dt.$$

My attempt: Using the identity, $(f-g)^+ = \frac{1}{2}( |f-g| + (f-g) )$, if the following equality of sets $$\{g(x) < t < f(x)\} = \{-\frac{1}{2}|f(x)-g(x)| < t < \frac{1}{2}(f(x)-g(x))\}$$ hold for all $t \in \mathbb{R}$, I am able to directly apply the layer-cake representation. However, I am only able to show the equality of sets holds for all $t\geq0$. Are there any other ways I can solve this problem? Any hints are appreciated.

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This is an application of Fubini's theorem to $$ I = \iint_{\Bbb R^d \times \Bbb R} h(x, t) \, d(x, t) $$ where the function $h: \Bbb R^d \times \Bbb R \to \Bbb R$ defined by $$ h(x, t) = \begin{cases} 1 & \text{ if } g(x) < t < f(x) \,.\\ 0 & \text { otherwise}. \end{cases} $$ For $x \in \Bbb R^d$ is $$ \int_{-\infty}^\infty h(x, t) \, dt = (f(x)-g(x))^+ \, , $$ and for $t \in \Bbb R$ is $$ \int_{\Bbb R^d} h(x, t) \, dx = |\{g(x) < t < f(x)\}| \, , $$ so that both the left-hand side and the right-hand side of your equation are equal to the integral $I$.

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