Finite-valued condition of measurable functions

Question

In page 28 of Real Analysis by Stein, it is stated that if $f$ is finite-valued then it is measurable iff the sets $\{a<f<b\}$ are measurable for every $a,b \in \mathbb{R}$. I cannot understand why the finite-valued condition is necessary. Because $\cup_{n=1}^{\infty} \{a<f<a+n\}=\{a<f\}$ for any $a\in \mathbb{R}$. For the reverse side, $\{f<b\}\cap \{a<f\}=\{a<f<b\}$.

Answer 1

The equation $\cup_{n=1}^{\infty} \{a<f<a+n\}=\{a<f\}$ is false if $f$ is allowed to take the value $\infty$. (Note that LHS is contained in $\{x:f(x)<\infty\}$). If $f\equiv \infty I_A -\infty I_{A^{c}}$ where $A$ is not measurable thne $\{a<f<b\}$ is empty, hence measurable for all $a,b$ but $f$ is not measurable because $f^{-1}(\{\infty\})=A$ is not measurable.