Is the mean value theorem of integral Calculus valid if f is not continuous?

Question

Suppose the real valued function f is Riemann integrable, but not continuous, over [a,b]. Then, does there exist some c in (a,b) such that f(c) = [Integral (a^b)f(x) dx]/(b -a)?

I tried to prove this based on the fact that the set of discontinuities of f in [a,b] has Lebesgue measue 0, but couldn't come to a conclusion yet.

Answer 1

No.

Consider the function f defined on [0, 1] by

f(x) = 0, if 0 <= x < 1/2
f(x) = 1, if 1/2 <= x <= 1.

Then the average value of f(x) on the interval [0, 1] is 1/2 (that is, [Integral (0^1)f(x)dx]/(1 - 0) = 1/2), but the function f does not assume the value 1/2 at any point in (0, 1).

---

I am pretty sure, however, that if a real valued function f is Riemann integrable on [a, b], and has the intermediate value property on [a, b] (that is, if f(x) < c < f(y) for x, y in [a, b], then there is a z between x and y such that f(z) = c), then the Mean Value Theorem does hold. (This is a more general condition than continuity; continuous functions always assume intermediate values, but functions assuming intermediate values are not necessarily continuous.)