Problem with "almost everywhere" concept.

I was given a problem, which was to show that two functions are equal almost everywhere, where one function is measurable and the other is not.

My problem is that the set where the equality fails is not measurable, although it is a subset of a set of measure zero (the measure space is not complete). Thus, I feel like I'm trying to prove a contradiction, because by definition, the set where the equality fails should be of measure zero. How do you show this?

I've looked around, and I read somewhere that the set where the equality fails doesn't have to be of measure zero, as long as it's a subset of a set of measure zero, but no proof was given. I'm assuming it was a modified version of the definition I was given. Intuitively, it makes sense ... but is there an actual proof for this?

P.S. In case you want more details, the measurable function was the constant function 0, and the non-measurable function was the characteristic/indicator function (where the value is 1 on the non-measurable set).

Re: Problem with "almost everywhere" concept.

Quote:

Originally Posted by

**Bingk** I was given a problem, which was to show that two functions are equal almost everywhere, where one function is measurable and the other is not.

My problem is that the set where the equality fails is not measurable, although it is a subset of a set of measure zero (the measure space is not complete). Thus, I feel like I'm trying to prove a contradiction, because by definition, the set where the equality fails should be of measure zero. How do you show this?

I've looked around, and I read somewhere that the set where the equality fails doesn't have to be of measure zero, as long as it's a subset of a set of measure zero, but no proof was given. I'm assuming it was a modified version of the definition I was given. Intuitively, it makes sense ... but is there an actual proof for this?

P.S. In case you want more details, the measurable function was the constant function 0, and the non-measurable function was the characteristic/indicator function (where the value is 1 on the non-measurable set).

I'm a little confused. Can you give a little more context about what measures/measure spaces we are looking at. Your counterexample doesn't work. If your consider the indicator function $\displaystyle \mathbf{1}_M$ then the fact that $\displaystyle \mathbf{1}_M$ agrees with $\displaystyle 0$ a.e. tells us that $\displaystyle M$ has measure zero, and is thus (if we are dealing with Lebesgue measure on $\displaystyle \mathbb{R}$ for example) measurable.

Re: Problem with "almost everywhere" concept.

Wikipedia :

Quote:

A nonmeasurable set is considered null if it is a subset of a null measurable set. Some references require a null set to be measurable; however, subsets of null sets are still negligible for measure-theoretic purposes.

(and I was told this way as well (Itwasntme))

@Drexel : the OP has a problem with the definition of null-measure, your sequence of propositions is not precise enough. Or so I guess.

Because you precisely can't talk about the measure of a non-measurable set. Even if its measure is globally 0, it's not correct to write the measure of this set.

But for our convenience, I think it's admitted that if a set is included in a set of null measure, then its measure is 0 (thanks to the sub-additivity of the measure, we have the right intuition)

Re: Problem with "almost everywhere" concept.

@Drexel: There was no specific measures/measure space, and in fact, the example you gave me is exactly why I'm having trouble, because the $\displaystyle M$ in your example is the non-measurable set that's a subset of a measure zero set. In any case, thank you! You basically confirmed my doubts of the validity of the question I was given.

@Moo: So, basically we can sort of abuse measurability by saying that non-measurable sets that are subsets of measure zero sets are also measure zero? (at least according to wiki and whoever told you)

Re: Problem with "almost everywhere" concept.

Quote:

Originally Posted by

**Bingk** I was given a problem, which was to show that two functions are equal almost everywhere, where one function is measurable and the other is not.

My problem is that the set where the equality fails is not measurable, although it is a subset of a set of measure zero (the measure space is not complete). Thus, I feel like I'm trying to prove a contradiction, because by definition, the set where the equality fails should be of measure zero. How do you show this?

I've looked around, and I read somewhere that the set where the equality fails doesn't have to be of measure zero, as long as it's a subset of a set of measure zero, but no proof was given. I'm assuming it was a modified version of the definition I was given. Intuitively, it makes sense ... but is there an actual proof for this?

P.S. In case you want more details, the measurable function was the constant function 0, and the non-measurable function was the characteristic/indicator function (where the value is 1 on the non-measurable set).

This is why you always pay close attention to the $\displaystyle \sigma$-algebra of subsets you're working with. For example the zero measure can be defined on any $\displaystyle \sigma$-algebra, but depending on the particular one you choose, even this trivial measure can have non-measurable subsets. This, of course, is artificial since you can always define in a natural way the zero measure on the power set of your measure space, and it can be proven that in a sense this can (almost) be done on every case, more formally we have

Suppose $\displaystyle (X,S,\mu)$ is a measure space then there exists a $\displaystyle \sigma$-algebra $\displaystyle S\subset S'$ and a measure $\displaystyle \mu'$ such that $\displaystyle \mu'|_S=\mu$ and $\displaystyle \mu'$ is complete.

This completness property is exactly what we want to have to be able to hand-wave a.e. differences, however not all measures have it: The Borel $\displaystyle \sigma$-algebra on $\displaystyle \mathbb{R}$ has $\displaystyle 2^{|\mathbb{N}|}$ elements, while there are $\displaystyle 2^{2^{|\mathbb{N}|}}$ subsets of Lebesgue measure zero. So Lebesgue measure restricted to the former is not complete, and it's one of the reasons we wanto to extend this measure.

Basically what you're being asked to prove is one of the reasons why we want our measures to be complete, they behave weirdly otherwise.

Re: Problem with "almost everywhere" concept.

Quote:

Originally Posted by

**Bingk** @Moo: So, basically we can sort of abuse measurability by saying that non-measurable sets that are subsets of measure zero sets are also measure zero? (at least according to wiki and whoever told you)

Yes. Also read Jose's pretty nice explanation :p

Re: Problem with "almost everywhere" concept.

Thanks! I was thinking about that also, extending the $\displaystyle \sigma$-algebra to be complete, and thus it would be measure zero.

I found out that our teacher gave us a different definition from what I've been reading, and it works for that (basically f=g a.e. if the complement of a set where it works has measure zero). But what he's been saying is what I've been reading, hehehe. Very sorry to cause all this trouble (Nod)

Thank you everyone! I love mathhelpforum.com :D