Show that a measurable function $f$ defined on $[0, 1]$ has the property for every $\epsilon > 0$ there is a $M_{\epsilon} > 0$ so that $λ({x ∈ [0, 1] : |f(x)| ≤ M_{\epsilon}}) ≥ 1 − \epsilon$ if and only if $f$ is finite almost everywhere.
I just wondering how to do this. Any ideas?
First let's introduce some notation to simplify things a bit. Let $F_n = \{x \in [0, 1] : |f(x)| \leq n\epsilon\}$ and put $E_n = \{x \in [0, 1] : |f(x)| > n\epsilon\}$.
Let $\epsilon > 0$ and assume that $M$ is such that $\lambda(F_M) \geq 1 − \epsilon$. Then by properties of measure we get $$\epsilon \geq 1 - \lambda(F_M) = \lambda([ 0,1 ]) - \lambda(F_M) = \lambda(E_M).$$ Letting $\epsilon \to 0$ we clearly have $M \to \infty$, giving that $\lambda(E_\infty) = 0$.
On the other hand, assume that $f$ is finite almost everywhere. Then $\lambda(E_\infty) = 0$. Let $\epsilon > 0$. Then (since $\lambda$ is continuous) there is an $M$ such that $\lambda(E_M) \leq \epsilon$. Thus $$1 - \epsilon \leq \lambda([0,1]) - \lambda(E_M) = \lambda(F_M).$$