# Can we say that $\lambda (A) = 0\$?

Let $$(X, \mathcal A, \mu)$$ be a $$\sigma$$-finite measure space and let $$f : X \longrightarrow [0,\infty]$$ be a non-negative $$\mu$$-measurable function on $$X.$$ Consider the set \begin{align*} A & := \left \{t \in [0,\infty]\ \bigg |\ \mu \left (\left \{x \in X\ \bigg |\ f(x) = t \right \} \right ) \gt 0 \right \}. \end{align*} Can we say that $$\lambda (A) = 0\$$?

where $$\lambda$$ denotes the Lebesgue measure on $$\Bbb R.$$

I think it is the case but can't able to prove it. Do anybody have any idea about that?

• Let $$C_0\subset C_1 \subset \ldots \subset X$$ be measurable subsets such that $$\mu(C_n) < \infty$$ for all $$n\in\mathbb N$$. Then, we have : $$A = \bigcup_{n\in\mathbb N}\bigg\{t\in[0,\infty]\bigg|\mu\big(\{x\in C_n|f(x) = t\}\big)>0\bigg\}$$ so wlog we can assume that $$\mu$$ is finite.
• The measurable sets $$\{x \in X |f(x) = t\}$$ form a partition of $$X$$ as $$t$$ ranges over $$[0,+\infty]$$. Therefore, since $$\mu$$ is finite, only countably many can have positive measure. Hence $$A$$ is countable and $$\lambda(A) = 0$$