Let $(\Omega, \mathcal{M}, \mu)$ be a measure space and let $f\ge 0$ be a measurable function on $\Omega$. Suppose that $f$ satisfies the following properties:
- For all $\varepsilon > 0$ there exists a $\delta > 0$ such that for any measurable subset $A\in \mathcal{M}$ such that $\mu(A)<\delta$, we have $$\int_A f(x)\, \mu(dx) < \varepsilon\;$$
- For all $\varepsilon > 0$ there exists a measurable subset $B\in \mathcal{M}$ such that $\mu(B)<\infty$ and $$\int_{\Omega \setminus B} f(x)\, \mu(dx) < \varepsilon.$$
Question Does it follow that $f\in L^1(\Omega)$?
Clearly it is sufficient to consider only the special case in which $\Omega$ is a probability space. Then I am able to answer affirmatively if $\Omega$ is non-atomic by appealing to the decomposition result explained in this post by Byron Schmuland. That is, since $\Omega$ supports a random variable with uniform $(0, 1)$-distribution, for a fixed value of $\varepsilon$ (say, $\varepsilon = 1$) we can decompose $\Omega$ into a finite disjoint union of sets of measure smaller then $\delta$ and then conclude by means of assumption 1.
But this seems too complicated for a result which I feel should be trivial. What am I overlooking?
Thank you.
Here's an idea for a proof which at first glance seems simpler:
If $A=\{x:f(x)=\infty\}$, then $\mu(A)=0$; else, 1. would be false. Thus, for the sets $A_n=\{x \mid f(x)>n\}$, we have $\lim\limits_{n\rightarrow\infty} \mu(A_n)=\mu(A)$. So, choose $N$ with $\mu (A_N)<\delta_1$ ...
But the implication in the first sentence above relies upon being able to find subsets of $A$ of arbitrarily small measure. This can be done using this result of Wacław Sierpiński. (I suspect, however, that this is exactly what you wanted to avoid.)