Except on a set of measure epsilon vs Almost Everywhere

[lugita15]

"Except on a set of measure epsilon" vs "Almost Everywhere"

There are certain results in analysis which say that a property P holds everywhere except on a set of arbitrarily small measure. In other words, for any epsilon you can find a set F of measure less than epsilon such that P holds everywhere except possibly on F. For convenience, I'll say that such a property P holds "almost almost everywhere." For instance, a measurable function is continuous almost almost everywhere, and a convergent sequence of measurable functions is uniformly convergent almost almost everywhere.

My question is, why aren't you allowed to go from "almost almost everywhere" to "almost everywhere"? Intuitively, this is how I'd expect things to work. For each k, let F_k be a set of measure less than 1/k, such that P holds everywhere except F_k. Then since the set Z of points on which P does not hold is a subset of F_k for each k, it follows that Z is a subset of the intersection of all the F_k. But clearly this intersection has measure zero, so Z has measure zero as well. Where is the error in my reasoning?

Any help would be greatly appreciated.

Thank You in Advance.

 

[Erland]

I think you misunderstood what these theorems you refer to actually say.
The theorem you mention about continuous functions does not say that for every measurable function f and every e > 0 , f is continuous everywhere outside a set of measure e, but it says that that given a measurable f, to every e > 0 there exists a continuous function f_e, which may depend upon e, such that f = f_e everywhere outside a set of measure e.
So, there are different continuous functions for different e(psilons), and there might not exist a continuous function that works for all epsilons.

 

[lugita15]

The theorem you mention about continuous functions does not say that for every measurable function f and every e > 0 , f is continuous everywhere outside a set of measure e, but it says that that given a measurable f, to every e > 0 there exists a continuous function f_e, which may depend upon e, such that f = f_e everywhere outside a set of measure e.

Well, that's not how the result is stated in Wheeden and Zygmund. It says that if f is a measurable function defined on a measurable set E, then for any epsilon > 0 there exists a closed subset F of E such that |E-F| < epsilon and f is continuous on F. That seems to be a clear case of "almost almost everywhere."

 

[Erland]

Well, that's not how the result is stated in Wheeden and Zygmund. It says that if f is a measurable function defined on a measurable set E, then for any epsilon > 0 there exists a closed subset F of E such that |E-F| < epsilon and f is continuous on F. That seems to be a clear case of "almost almost everywhere."

For a moment, I thought I got confused (due to age, perhaps ).

However, there are subtleties here making it easy to go astray. In the theorem you mention "f is continuous on F" must be interpereted as "the restriction of f to F is continuous on F". It cannot be interpreted as "the original function f [being a function defined on E] is continuous at every point of F", because with the latter interpretation, the statement of the theorem is false for e.g. the function f defined on the interval [0,1] whose value is 0 for all rational numbers in the interval and 1 for all irrational numbers in the interval. This function is continuous nowhere in the interval, but the restriction to any subset of the irrationals in the interval is continuous. Since the set of rationals in the interval has measure 0, the conclusion of the theorem is true with the first interpretation.

So, we still have different functions for different epsilons, namely different restrictions of the original function.

Although it is so in the given example, it is not obviuos that there always must be a set F whose complement has measure zero such that the restriction of f to F is continuous on F. I am quite sure that it needs not be so, because otherwise someone would have proved such a theorem, but I cannot find any counterexample. Perhaps some of you people here know of one?

See http://en.wikipedia.org/wiki/Lusin's_theorem.

 

[lugita15]

For a moment, I thought I got confused (due to age, perhaps ).

However, there are subtleties here making it easy to go astray. In the theorem you mention "f is continuous on F" must be interpereted as "the restriction of f to F is continuous on F". It cannot be interpreted as "the original function f [being a function defined on E] is continuous at every point of F", because with the latter interpretation, the statement of the theorem is false for e.g. the function f defined on the interval [0,1] whose value is 0 for all rational numbers in the interval and 1 for all irrational numbers in the interval. This function is continuous nowhere in the interval, but the restriction to any subset of the irrationals in the interval is continuous. Since the set of rationals in the interval has measure 0, the conclusion of the theorem is true with the first interpretation.

So, we still have different functions for different epsilons, namely different restrictions of the original function.

Although it is so in the given example, it is not obviuos that there always must be a set F whose complement has measure zero such that the restriction of f to F is continuous on F. I am quite sure that it needs not be so, because otherwise someone would have proved such a theorem, but I cannot find any counterexample. Perhaps some of you people here know of one?

See http://en.wikipedia.org/wiki/Lusin's_theorem.

OK, so the result is f is continuous relative to F, which is different from saying that it's continuous on the points is F.

So now my question is, if F_{1}\subset F_{2}\subset ... and f is continuous relative to each F_{k}, then is f continuous relative to \bigcup F_{k}?

 

[Erland]

So now my question is, if F_{1}\subset F_{2}\subset ... and f is continuous relative to each F_{k}, then is f continuous relative to \bigcup F_{k}?

No, and here it is much easier to find a counterexample. Let f(0)=0 and f(x)=1 for x>0. Let F_k=\{0\}\cup[1/k,1]. Then f is continuous relative to each F_k, but not relative to the union, which is [0,1], where 0 is a point of discontinuity.

That dosen't seem to help us to construct a counterexample in the situation in the previous posts, though.

 

[lugita15]

No, and here it is much easier to find a counterexample. Let f(0)=0 and f(x)=1 for x>0. Let F_k=\{0\}\cup[1/k,1]. Then f is continuous relative to each F_k, but not relative to the union, which is [0,1], where 0 is a point of discontinuity.

That dosen't seem to help us to construct a counterexample in the situation in the previous posts, though.

OK, so that's where my argument breaks down. Now what about the other result I mentioned? If again F_{1}\subset F_{2}..., and \left\{f_{n}\right\} is a sequence of functions which converges uniformly to f on each F_{k}, then does \left\{f_{n}\right\} converge uniformly to f on \bigcup F_{k}?

 

[Office_Shredder]

Still not true. Consider the functions on [0,1] with f_k(1/k)=1, f_k(x)=0 if x>2/k, and between 0 and 2/k f_k forms a peak by increasing to 1 linearly then decreasing to 0. On each [1/n,1] the sequence converges uniformly to 0, but on the union, (0,1] each f_k has a value of x for which f_k(x)=1