How to proof $ \int_{0}^{x}\int_{0}^{s}d\ f_{+}^{'}(t)\ d\ s = \int_{0}^{x}(x-t)_{+}d\ f_{+}^{'}(t) $?

Question

How to proof the following equation, $$ \int_{0}^{x}\int_{0}^{s}d\ f_{+}^{'}(t)\ d\ s = \int_{0}^{x}(x-t)_{+}d\ f_{+}^{'}(t). $$ Here, $(x-b)_{+} = max{x-b, 0}$; $f_{+}^{'}$ denotes the right derivative of $f$.

Answer 1

I am guessing that $g_+$ means $\max\{g,0\}$. Let $\chi_E$ be the indicator function of a set $E$. If $P$ is the subset of $[0,x]$ on which $f'>0$, then $$\int_0^sdf'_+(t)=\int_0^x\chi_P(t)\ \chi_{[0,s]}(t)\ df'(t).$$ By Fubini's theorem, we get $$\int_0^x\int_0^s df_+'(t)\ ds=\int_0^x\int_0^x\chi_P(t)\ \chi_{[0,s]}(t)\ df'(t)\ ds=\int_0^x\int_0^x \chi_{[0,s]}(t)\ ds\ \chi_P(t)\ df'(t).$$ Therefore, $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x\int_0^x\chi_{[t,x]}(s)\ ds\ \chi_P(t)\ df'(t)=\int_0^x\int_t^x ds\ \chi_P(t)\ df'(t).$$ Consequently, $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x(x-t)\ \chi_P(t)\ df'(t)=\int_0^x(x-t)\ df'_+(t).$$ But as $x\geq t$ for all $t\in[0,x]$, $(x-t)=(x-t)_+$, and $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x(x-t)_+\ df'_+(t).$$

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Edit after OP's correction

If $P$ is the subset of $[0,x]$ on which $f'>0$, then $$\int_0^sdf'_+(t)=\int_0^x\chi_{[0,s]}(t)\ df'_+(t).$$ By Fubini's theorem, we get $$\int_0^x\int_0^s df_+'(t)\ ds=\int_0^x\int_0^x\chi_{[0,s]}(t)\ df_+'(t)\ ds=\int_0^x\int_0^x \chi_{[0,s]}(t)\ ds\ df_+'(t).$$ Therefore, $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x\int_0^x\chi_{[t,x]}(s)\ ds\ df'_+(t)=\int_0^x\int_t^x ds\ df_+'(t).$$ Consequently, $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x(x-t)\ df'_+(t).$$ But as $x\geq t$ for all $t\in[0,x]$, $(x-t)=(x-t)_+$, and $$\int_0^x\int_0^s df'_+(t)\ ds=\int_0^x(x-t)_+\ df'_+(t)=\int_0^\infty(x-t)_+\ df'_+(t).$$