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For example, for the function $$ f(x)=|x|,\quad\text{$-1\le x\le1$,} $$ the subdifferential $D^{+}f(x)$ is $$ D^{+}f(x)=\begin{cases} -1\quad&\text{for $x\in[-1,0)$,}\\ [-1,1]\quad&\text{at $x=0$,}\\ 1\quad&\text{for $x\in(0,1]$.} \end{cases} $$

I now want to know what the the following integral is: (formally) $$ \int_{-1}^{1}D^{+}f(x)dx=?. $$ However, as you see, $D^{+}f$ is multi-valued function. I thought that it is alright to interpret this integral as Lebersgue integral since we may ignore the point $x=0$, that is, $$ \int_{-1}^{1}D^{+}f(x)dx=-1\cdot|[-1,0)|+1\cdot|(0,1]|=0, $$ but what happend if we interpret in the sense of Riemann integral? Can we interpret in this sense in the first place? I have searched but I didn't find any information.

Please give me some comments for my question and my interpretation in the sense of Lebesgue integral if you know.

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Source: Proposition 1.6.1 from Convex functions and their applications: a contemporary approach by Niculescu and Persson.

Let $f:I\to\mathbb{R}$ be a continuous convex function and let $\varphi:I\to\mathbb{R}$ be a function such that $\varphi(x)\in\partial f(x)$ for every $x\in\operatorname{int} I$. Then for every $a<b$ in $I$ we have $$ f(b)-f(a)=\int^b_a\varphi(t)dt.$$

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  • $\begingroup$ Does such a result where $I$ is replaced by a Hilbert space? $\endgroup$ – AIM_BLB Feb 4 at 7:06

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