# If $\int|f|^p\ d\mu=K$, then $\int \lvert f \rvert^p\ \chi_E\ d\mu\le K\mu(E)$

Let $$f\in L_p$$, so

$$\lVert f \rVert_p <+\infty \Longrightarrow \int\vert f\rvert^p\ d\mu=K<\infty$$

For a demonstration of an exercise I assumed that,

$$\int \lvert f \rvert^p\ \chi_E\ d\mu\le\int K\chi_E\ d\mu=K\mu(E)$$

But a colleague questioned me about that statement, and I ended up not so sure about it anymore, because it’s an old proof and I don’t remember exactly what I thought when I wrote this. I was unable to demonstrate this statement or find a counterexample.

So, my question is whether the above implication is true or not.

• Try it with $p=1$ and an $f$ that takes two different positive values. – Robert Israel Oct 28 at 18:27

The inequality can be written as $$\frac{1}{\mu(E)}\int_E|f|\le\|f\|$$ (Put $$|f|^p$$ instead, if you wish.)
That the average of a function on a subset need not be less than the total function (or even the total average) is obvious: Take $$f(x)=\chi_{[0,1/2)}+2\chi_{[1/2,1)}$$ and $$E=[1/2,1)$$, $$2>\frac{2+1}{2}$$
No it's not. Consider for example $$f(x)=\sum_{n=1}^{\infty} n \chi_{[\frac{1}{n^3+1}, \frac{1}{n^3})}$$ for $$x\in (0,1)$$ and $$f\equiv 0$$ elsewhere.
Then $$f$$ is in $$L^1$$, and lets say it has norm $$\infty>K>0$$. Choose $$n$$ large enough so that $$n>K$$. Then for $$E=[\frac{1}{n^3+1}, \frac{1}{n^3})$$ you have that $$\int_E f > \int_E K$$