If $\int_X f \,d\mu = \infty$ then for each $C > 0$ there exists a set $E$ of finite measure such that $\int_E f \,d\mu > C$

Question

Let $(X, \mathcal{M}, \mu)$ be a semifinite [definition below] measure space and let $f: X \to [-\infty, \infty]$ be an extended $\mu$-integrable function such that $\int_X f \,d\mu = \infty$. Does the following condition hold?

For each $C > 0$ there exists a set $E \in \mathcal{M}$ with $\mu(E) < \infty$ such that $\int_E f \,d\mu > C$.

Clearly, it suffices to consider the case where $f$ is nonnegative since we must have $\int_X f^{+} \,d\mu = \infty$ and $\int_X f^{-} \,d\mu \in \mathbb{R}$, where $f^{+} := \max(f,0), f^{-} := - \min(f, 0)$, and $f = f^{+} - f^{-}$.

If $(X, \mathcal{M}, \mu)$ is $\sigma$-finite, then the claim is easy to prove: Assuming $f \geq 0$, take a measurable partition $\{E_k\}_{k=1}^{\infty}$ of $X$ such that $\mu(E_k) < \infty$ for all $k$, and set $X_n := \bigcup_{k=1}^{n} E_k$. Then $X_n \nearrow X$, $\mu(X_n) < \infty$, and $\lim\limits_{n \to \infty} \int_{X_n} f \,d\mu = \lim\limits_{n \to \infty} \int_X 1_{X_n} f \,d\mu = \infty$ by the Monotone Convergence Theorem. Consequently, for each $C > 0$ there exists some $n \in \mathbb{N}$ such that $\int_{X_n} f\,d\mu > C$. But what about the semifinite case? I'm not sure how to proceed without the $\sigma$-finite assumption.

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This is my definition of semifinite:

Definition. A measure space $(X, \mathcal{M}, \mu)$ is called semifinite if, for every set $A \in \mathcal{M}$ with $\mu(A) = \infty$, there exists a set $E \in \mathcal{M}$ such that $E \subset A$ and $0 < \mu(E) < \infty$.

Update 1/22/25: I think I have it now, thanks to the following lemma.

Lemma (Exercise 1.14 of Folland, p.27) If $\mu$ is a semifinite measure on $(X,\mathcal{M})$ and $\mu(A) = \infty$, then for any $C > 0$ there exists $E \in \mathcal{M}$ with $E \subset A$ such that $C < \mu(E) < \infty$.

My proof. Assume $f$ is nonnegative and $\int_X f \,d\mu = \infty$. Then there there exists an increasing sequence of nonnegative simple functions $(\phi_n)_{n=1}^{\infty}$ such that \begin{align*} \int f \,d\mu = \lim_{n \to \infty} \int \phi_n \,d\mu. \end{align*}

Let $C > 0$. Then there exists some $N \in \mathbb{N}$ such that $\int_X \phi_N \,d\mu > C$. Since $\phi_N$ is simple, it has the form $\phi_N = \sum_{j=1}^{m} a_j 1_{A_j}$, where the $a_j$'s are positive real numbers and the $A_j$'s are measurable sets. If $\mu(A_j) < \infty$ for all $j$, then $E := \bigcup_{j=1}^{m} A_j$ has finite measure and \begin{align*} \int_E f \,d\mu \geq \int_E \phi_N \,d\mu = \int_X \phi_N \,d\mu > C \end{align*} and we're done. On the other hand, if $\mu(A_j) = \infty$ for some $j$, then by the lemma there exists some set $E \in \mathcal{M}$ such that $E \subset A_j$ and $\frac{C}{a_j} < \mu(E) < \infty$, which then implies \begin{align*} \int_E f \,d\mu \geq \int_E \phi_N \,d\mu \geq \int_E a_j 1_{A_j} \,d\mu = a_j \mu(E) > C. \end{align*}

Is this a sound proof?

Answer 1

New answer with the correct definition of semifinite measure

If we rephrase your problem, you are asking to show that $$ S:=\sup\left\{\int_E\vert f\vert\,d\mu\mid E\in\mathcal M,\ \mu(E)<+\infty\right\}=+\infty. $$

Suppose then that $S<+\infty$, and let us show this implies a contradiction.

By definition of the supremum, there exists a nondecreasing sequence $(E_n)_{n\in\mathbb N}\in\mathcal M^{\mathbb N}$ such that $\mu(E_n)<+\infty$ for all $n\in\mathbb N$ and $\int_{E_n}\vert f\vert\,d\mu\to S$ as $n\to+\infty$. Let then $$ E=\{f=0\}\cup\bigcup_{n\in\mathbb N}E_n, $$ so that $\int_E\vert f\vert\,d\mu=S<+\infty$. As $\int_X\vert f\vert\,d\mu=+\infty$, we have that $\int_{X\backslash E}\vert f\vert\,d\mu=+\infty$. And because $S<+\infty$, we necessarily have $\mu(X\backslash E)=+\infty$. As $\mu$ is semifinite, there exists $F\in\mathcal M$ such that $F\subset X\backslash E\subset\{\vert f\vert>0\}$ and $0<\mu(F)<+\infty$. This implies that $$ \varepsilon:=\int_F\vert f\vert\,d\mu>0. $$

Therefore, for all $n\in\mathbb N$, $\mu(F\cup E_n)<+\infty$ and $$ \int_{F\cup E_n}\vert f\vert\,d\mu\to S+\varepsilon>S, $$ which contradicts the definition of $S$.

Old answer which applies to s-finite measures but not semifinite.

No.

For instance for $X=\{0\}$ and $\mathcal M=\{\emptyset,\{0\}\}$, then the measure $\mu=\sum_{n\in\mathbb N}\delta_0$ is semifinite.

Then any nonzero function $f:X\to[-\infty,+\infty]$ is measurable and satisfies $\int_X\vert f\vert\,d\mu=+\infty$. But for $E\in\mathcal M$, $\mu(E)<+\infty$ iff $E=\emptyset$, which implies $\int_E\vert f\vert\,d\mu=0$.