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If a function $f$ is continuous in $[0,\Delta]$ it is pretty easy to prove that

$$ \exists c\in(0,\Delta):\frac{1}{\Delta}\,\int_{0}^{\Delta}f(t)dt=f(c) $$

It is enough to apply Lagrange's to the function $F(t)=\int_0^tf(s)ds$. Is it possible to derive the same result assuming only that $f$ is Riemann-integrable in $[0,\Delta]$ ?

AlmostSureUser
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1 Answers1

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No. For example consider the Riemann integrable function $f:[0,1]\rightarrow \{0,1\}$ defined by

$$f(x)= \begin{cases} 0, & 0\leq x\leq \frac{1}{2}\\ 1, & \frac{1}{2}<x\leq 1. \end{cases}$$

Then $\int_{0}^{1}f(x)dx=\frac{1}{2}$ (For this example $\Delta=1$) which is not equals to $f(c)$ for any $c\in[0,1]$.

Surajit
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