Mean value theorem for Lebesgue integral: $\int_a^bf(t)dt=(b-a)f(c)$ for some $c$.

Question

Let $f:[a;b]\to\mathbb R$ be any measurable function respect to Lebesgue measure. I wish to obtain the following: There exists $c$ belongs to the interval $[a;b]$ $\int_a^bf(t)dt=(b-a)f(c)$.

So my question is that for which function $f$, we shall have the above equality? Is that true for any Lebesgue measurable function?

Answer 1

Define $f$ on $[-1,1]$ by $f(x)=-1$ if $x\leqslant0$ and $f(x)=1$ if $x\gt0$. Then $\int\limits_{-1}^1f(t)\mathrm dt=0$ hence $f$ does not satisfy the first mean value theorem for integration.

Thus the result relies crucially on the fact that the function $f$ satisfies the intermediate value theorem (these are the so-called Darboux functions), in particular continuous functions are allright.

Answer 2

I think that it would be more correct if the statement sounded a little differently:

$$\int\limits_{a}^{b}f(x)\,\mathrm dx=(b-a)C, \text{ where } C \in [\inf\{f(x),x\in [a,b]\},\sup \{f(x),x\in [a,b]\}] .$$

This statement holds for a wider class of functions.In previous formulation the statement was true only for continuous functions. Moreover, the Lebesgue integrals are equal for equivalent functions, but if the functions are equivalent, one of them can have points of discontinuity, which makes this statement have no sense.