# Equi-integrability of a single function: is it the same as summability?

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:

1. 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 $$\int_A f(x)\, \mu(dx) < \varepsilon\;$$
2. 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.

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If $A=\{ x : f(x)=\infty\}$ then $\mu(A)=0$; else, 1. would be false. If $A_n=\{x | f(x)>n\}$, then $\lim_n \mu(A_n)=\mu(A)$. So choose $N$ with $\mu(A_n)<\delta_1$... But this relies upon finding subsets of $A$ with arbitrarily small measure. This can be done, according to this result. I suspect, however, that this is exactly what you wanted to avoid. – David Mitra Dec 15 '12 at 22:59
@DavidMitra: Yes, exactly. The post by Byron Scmhuland I am referring to uses a probabilistic language to say the same thing as that Wikipedia article. Thank you very much for your comment! I was afraid that the answer was absolutely trivial and that I was too dumb to see it. – Giuseppe Negro Dec 15 '12 at 23:05
@DavideGiraudo Sorry, apparently I did ... I'll post it shortly. – David Mitra Feb 17 '13 at 21:16

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.)