lesbesgue integral intermediate value

Question

Let $f:K\subset R^n \rightarrow R$ be a continous function and $K$ a compact set

Proof the following statement about the lebesgue integral.

$$\int_K fd\lambda =\lambda(K)f(\epsilon)$$ for some $\epsilon \in K.$

I am stuck with this problem and would appreciate a hint

Answer 1

$\lambda(K)\min f\leq\displaystyle\int_{K}fd\lambda\leq\lambda(K)\max f$, so by $\min f\leq\dfrac{1}{\lambda(K)}\displaystyle\int_{K}fd\lambda\leq\max f$, by Intermediate Value Theorem, there is some $\epsilon\in[m,M]$ such that $f(\epsilon)=\dfrac{1}{\lambda(K)}\displaystyle\int_{K}fd\lambda$, here $f(m)=\min f$, $f(M)=\max f$.

If $\lambda(K)=0$, then the integral is zero, and $\epsilon$ can be taken arbitrarily.

The above proof is valid for one-dimensional and when $K$ is connected. For higher dimension, there is no such $[m,M]$, but we still can make use of Intermediate Value Theorem if $K$ is connected.

Answer 2

This is false without assuming something about $K$, for example that $K$ is connected. For example, let $K=[0,1]\cup[2,3]\subset\mathbb R$ and let $f=0$ on $[0,1]$, $f=1$ on $[2,3]$. Then $\int_K f=\frac12\lambda(K)$, but there is no $\epsilon$ with $f(\epsilon)=\frac12$. (Or maybe you meant to say $f:\mathbb R^n\to \mathbb R$?)