Bounded function in a finite measurable space.

Question

Let $(X,\mathcal{A},\mu)$ a finite measurable space. Let $f:X\to\mathbb{R}$ a $\mathcal{A}$-measurable function. Show that for all $\epsilon>0$ exist $A\in \mathcal{A}$ so that $f|_{A}$ is bounded and $\mu(A^c)<\epsilon$

Answer 1

Let $X_n = f^{-1} (-n, n)$. Then note that $X_n \subset X_{n + 1}$ for every $n \in \mathbb{N}$.

Also $\bigcup X_n = X$.

By continuity of measure, $\mu (X_n) \to \mu (X)$.

As $\mu (X) < + \infty$, for every $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that whenever $n \ge N$, $\mu (X) - \mu (X_n) < \varepsilon$.

Therefore $\mu (X_N^c) = \mu (X) - \mu (X_N) < \varepsilon$.

Also $f$ is bounded on $X_N$ by definition.