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I notice a discrepancy between standard assumption of MVT for derivatives and for integrals. The former does not requires that the differential to be continuous whereas the latter requires the function (integrand) to be continuous. That is, if we interpret the statement $f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{\int_a^bf'(x)dx}{b-a}$ as integral MVT, we notice that the two versions of MVT is equivalent if we assume $f'(x)$ is continuous--which is not required for derivative MVT. This seems odd to me. Why would integral MVT need stronger assumption here? What is the weakest assumption for integral MVT?

Daniel Li
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It's not really the same theorem, but derivatives of a differentiable function have the Darboux property, even when they are discontinuous. As long as you can make sense of the integral, $ (b-a)^{-1} \int_a^b f'(x)\; dx$ must be between the least and greatest values of $f'$ on $[a,b]$, and Darboux says this is $f'(c)$ for some $c$. The problem is to make sense of the integral. Are you using Riemann or Lebesgue integration?

For example, see this answer for a function that is differentiable but not absolutely continuous on $[-1,1]$. In this case $\int_{-1}^1 f'(x)\; dx$ does not exist, even as a Lebesgue integral.

Robert Israel
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