For an integrable function $f$, the measure of the set $\{x: |f(x)|=a \}$ is zero for all but countably many $a's$ Suppose $(\Omega, \mathcal{F}, \mu)$ is a probability space and $f$ is integrable function. Then, is it true that $\mu \{ \omega: |f(w)|=\alpha\}=0$ for all but countably many $\alpha$ ? \
 A: It doesn't depend on integrability of $f$. Measurability suffices.
Let 
$$ A_n = \{ a\in [0,+\infty] : \mu(|f|=a)>1/n \}$$
Then the cardinality of $A_n$ is at most $n$.
And $\cup_{n\geq 1} A_n$ is the set of $a$'s for which $\mu(|f|=a)$ is strictly positive.
Remark: The result  also holds for any $\sigma$-finite measure space.
A: Suppose that there exists an uncountable set $A$ such that, if $a\in A$, then $\mu\left(\{\omega\in\Omega\Big{|}|f(\omega)|=a \}\right)>0$. For every $n\in\mathbb N$, set $A_n$ to be the set of $a\in A$ such that $$\frac{1}{n}\leq\mu\left(\{\omega\in\Omega\Big{|}|f(\omega)|=a \}\right)<\frac{1}{n-1},$$ then $\bigcup_{n\in\mathbb N}A_n=A$. Since the $A_n$ are countably many and $A$ is uncountable, there exists $N\in\mathbb N$ such that $A_N$ is uncountable. Therefore, we can find $\varepsilon>0$ and a sequence $(a_m)$ in $A_N$, such that $a_m>\varepsilon$ for all $m\in\mathbb N$.
Set $$B_m=\{\omega\in\Omega\Big{|}|f(\omega)|=a_m \}.$$ We then compute $$\int_X|f|\,d\mu\geq\sum_{m\in\mathbb N}\int_{B_m}|f|\,d\mu=\sum_{m\in\mathbb N}\int_{B_m}a_m\,d\mu>\sum_{m\in\mathbb N}\varepsilon\mu(B_m)\geq\sum_{m\in\mathbb N}\frac{\varepsilon}{N}=\infty,$$ which is a contradiction.
A: Hint: Let $X$ be a measure space such that $\mu(X)>0$, consider $Y=\bigcup_{a\in R}X_a$ where $(X_a,\mu_a)$ is isomorphic to $\mu_a$, consider $f:Y\rightarrow R$ whose restriction to $X_a$ is the constant function $a$.
