I have the following problem that I'm trying to puzzle through.
Let $\mu$ be a positive measure on $[0,1]$ with $\mu([0,1])=1$ and let $f\geq 0$ be a summable function on $[0,1]$. Suppose that $\int_{[0,1]}f~d\mu<\frac{1}{10}$. Define $X=\{x\in[0,1]~:~f(x)<1\}$. Prove that $\mu(X)\geq \frac{9}{10}$.
I know that since $\int_{[0,1]}f~d\mu$ is the supremum of all integrals $\int_{[0,1]}\psi~d\mu$, where $\psi$ is simple and $0\leq\psi\leq f$, this supremum must be less than $\frac{1}{10}$.
I was thinking that maybe I need to use the characteristic function of the set $X$ and construct some sort of sequence of simple functions to approximate $f$. But I'm not really sure if this is the right approach, or what I would need to do in this case. Any input is appreciated!
Hint. $$\mu(X^c) =\int_{X^c}1\ d\mu \leq \int_{X^c} f \ d\mu \leq \int_{[0,1]}f\ d\mu< \frac 1 {10}$$ Why do these inequalities hold? Finally, what is the relation between $\mu(X^c)$ and $\mu(X)$?