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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?

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1 Answer 1

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  • 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 $\lambda(A) = 0$

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