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

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?

• I think the existence of $c$ cannot be simplified much further... (This is true for continuous functions. It can be false for many Riemann integrable functions.) Aug 1, 2013 at 9:04
• Yes I think continuity is essential but not sure how to prove it. As for checking the falsity for Riemann integrable functions, one just needs to first get continuous functions $f$ such that a finite number of $c$ are available and then change value of $f(c)$ at these finite number of points $c$. The new function is still Riemann integrable, but not continuous and now there is no $c$ satisfying mean value theorem. Aug 1, 2013 at 9:49
• Got something from an answer below?
– Did
Aug 13, 2013 at 8:53

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.

• Not much Lebesgue in here, this is all very much Riemann-Darboux-Cauchy.
– Did
Aug 1, 2013 at 10:26

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.

• "In previous formulation the statement was true only for continuous functions"... Nope.
– Did
Aug 1, 2013 at 13:03
• I meant the authors' of the thread statement. If the function is continous and the integral is Reimans', then the first mean value theorem holds.
– cool
Aug 1, 2013 at 14:08
• OK. I misunderstood.
– Did
Aug 1, 2013 at 15:23