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Suppose a property $P$ may or may not hold on a measurable subset of a measure space with measure $\mu$. Is

for every $ε > 0$, there exists a measurable subset $B$ such that $μ(B) < ε$, and $P$ holds on $B^c$.

equivalent to

there exists a measurable subset $B$ with $μ(B) =0$, and $P$ holds on $B^c$?

For example, Egorov's theorem say that

if $(f_n)$ converges $μ$-almost everywhere on $A$ to a limit function $f$, where $A$ is a measurable subset of finite $μ$-measure, then for every $ε > 0$, there exists a measurable subset $B$ of $A$ such that $μ(B) < ε$, and $(f_n)$ converges to $f$ uniformly on the relative complement $A \setminus B$.

Thanks!

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  • $\begingroup$ Hint: Construct measurable subsets $B_n$ such that for any $n\in\mathbb{N}$, $\mu(B_n) = 1/n$ and $P$ holds on $B_n^c$. $\endgroup$ – Anon Jan 20 '13 at 23:46
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No, the two conditions are not equivalent. The second one, trivially, implies the first. But, if all you know is that for every $\epsilon>0$ there is a set with $\mu(B_\epsilon)<\epsilon$ such that property $P$ holds outside of $B_\epsilon$ you can't conclude $P$ holds almost everywhere. Take for instance the property $P$ to be "$\inf {|x|}$ is positive". Then, for every $\epsilon >0$ you can take $B_\epsilon = (-\epsilon/2,\epsilon/2)$ and then $P$ holds outside of $B_\epsilon$. But clearly $P$ does not hold almost everywhere.

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  • $\begingroup$ I am (almost!) surely misreading your example, but doesn't $P$ hold on $\mathbb R \setminus \{0\}$. $\endgroup$ – cardinal Jan 21 '13 at 0:01
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    $\begingroup$ thanks @cardinal, I corrected it. $\endgroup$ – Ittay Weiss Jan 21 '13 at 0:06
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No. The question Nearly, but not almost, continuous is about a version of this question for Lusin's theorem.

The same example given there can be used in a counterexample for Egoroff's theorem. If $C$ is a fat Cantor set and $g:[0,1]\to\mathbb R$ is defined by $g(x)=$ the distance from $x$ to $C$, let $f_n(x)=(1-g(x))^n$ for each $n\in\mathbb N$ (example borrowed from another question). Then $(f_n)_n$ converges pointwise (monotonically) to the characteristic function $f$ of $C$. Each $f_n$ is continuous, so if $(f_n)$ converges uniformly on $A\subseteq [0,1]$, then $f$ is continuous on $A$. But $f$ is not continuous on any subset of $[0,1]$ with measure $1$, as shown in the first linked question above.

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This might be a bit confusing in light of the (correct) other answers, but I would say the answer is yes. Suppose that for each $\epsilon>0$, there is a measurable set $N_\epsilon$ with measure less than $\epsilon$ such that property $P$ holds for every element of $N_\epsilon^C$. For each natural number $n>0$, let $N_{1/n}$ be of this form. Then an element that is not in $N_{1/n}$ for at least one $n$ has property $P$. Hence, an element not in $N=\bigcap_n N_{1/n}$ has property $P$ and $N$ has measure zero.

The reason that this argument does not work with, say, Egorov's theorem is that that result is not about a property of points in the measure space. It is about how the points in a set relate to each other, for that is what uniform convergence means, it is not a pointwise property. So one has to draw a distinction between properties of points and properties of measurable sets.

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