$\inf\{f(x) : x \in E \in \mathcal{M}, \mu(E) = \infty\} = 0$ Let $(X , \mathcal{M} , \mu)$ a measure space and let $f : X \to [0 , + \infty]$ be a function such that $\int_X f d \mu < \infty$. Then I have to show that
$$
\inf\{f(x) : x \in E \in \mathcal{M}, \mu(E) = \infty\} = 0.
$$
My attempt is the next: if $\int_X f d \mu < \infty$, then $A = \{x \in X : f(x) > 0\}$ can be written as
$$
A = \bigcup_{n = 1}^{\infty} A_n
$$
where
$$
A_n = \left\{x \in X : f(x) > \frac 1 n\right\}
$$
has finite mesure $\mu(A_n) \leq n \int_X f d \mu < \infty$. I think it's enough if we show that $A^c \neq \emptyset$. Is my argument helpful to show it? If $\mu(X) < \infty$, is $\{f(x) : x \in E \in \mathcal{M}, \mu(E) = \infty\} = \emptyset$?
 A: To avoid answering in the comments I've typed everything up here:
$\newcommand{\d}{\,\mathrm{d}}$Let $c$ equal the given infimum. As $f$ has nonnegative image, $c\ge0$. If $\mu(X)\lt\infty$ then the given question doesn't make sense, so suppose that this infimum is well defined; infimums always exist, so $c\in\Bbb R^+$ or $c=\infty$. If $c\ge1$, set $\alpha:=1$, and if $0\le c\le1$ put $\alpha:=c$.
Let $\varphi=\alpha\cdot\chi_E$ for some $E\in\mathcal{M}$ with $\mu(E)=\infty$ - we are assuming at least one $E$ exists. Then $\varphi\le f$ everywhere, since $\varphi=0\le f$ on $X\setminus E$ and $\varphi=\alpha\le c\le f$ on $E$.
Then by basic monotonicity of the Lebesgue integral:
$$\int_X\varphi\d\mu\le\int_Xf\d\mu$$
We are told that $\int_Xf\d\mu\lt\infty$. Suppose that $c\neq0$; then $\alpha\gt0$, and it follows from definition of the Lebesgue integral that:
$$\int_Xf\d\mu\ge\int_X\varphi\d\mu=\alpha\cdot\mu(E)=\infty,\,\alpha\gt0$$
Which is a contradiction; it then follows that $c=0$ since $c\neq0$ is  impossible, and $c\in\overline{\Bbb R}$.
