If I take pre-images of an increasing subset of the image, do their measures converge to that of the range?

Question

Consider an arbitrary function $f:[0,1]\to\mathbb{R}_+$ and consider the sequence $$m^n=\mu(\{x\mid f(x)\leq n\}),$$ where $\mu$ is the Lebesgue measure. In essence, $m^n$ measures the pre-image of $[0,n]$, and as such is trivially nondecreasing and bounded above by $1$. The monotone convergence theorem implies that this sequence converges to some $m=\lim_{n\to\infty} m_n$. However, intuitively I would say this bound needs to be $1$ (since I include more and more of the image), but is this true? Can anybody either provide a proof or a counterexample, or direct me to duplicate questions?

(I've tried searching the site, but I really do not know what keywords to look for... which is also the reason for the clumsy title.)

Answer 1

If $A_1,A_2,\dots$ are measurable sets with $A_1\subseteq A_2\subseteq\cdots$ and $A=\bigcup_{n=1}^{\infty}A_n$ then $\mu A_n\uparrow\mu A$ for (any) measure $\mu$.

I hope that's clear to you. If not then let me know.

Here $A_n:=f^{-1}((-\infty,n])$ and $A=[0,1]$. So...

Answer 2

First of all, $f$ needs to be measurable in order to talk about $\mu(\{x\in[0,1]:f(x)\leq n\})$. In that case the sets $\{x\in[0,1]:f(x)\leq n\}$ increase to $[0,1]$ and from properties of measure (continuity from below) $\lim_{n\to\infty} m_n=1$.