Measure Theory

[johnson123]

Problem: $$f_{n}\rightarrow f$$ in measure, $$\mu(\left\{f_{n}>h\right\})\leq A$$

Prove that $$\mu(\left\{f>h\right\})\leq A$$.

My Work:

Suppose not, then $$\mu(\left\{f>h\right\}) > A$$.

From the triangle inequality for measures we get

$$\mu(\left\{f>h\right\}) = \mu(\left\{f-f_{n}+f_{n}>h\right\}) \leq\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\})$$.

So $$A<\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\}) \leq\mu(\left\{f-f_{n}>0\right\}) + A$$.

Taking limits on both sides (n->00)yields:

$$A < 0 + A \Rightarrow\Leftarrow$$.

I do not have much of a background for analysis, so any suggestions are welcome.

 

[jostpuur]

So $$A<\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\}) \leq\mu(\left\{f-f_{n}>0\right\}) + A$$.

Taking limits on both sides (n->00)yields:

$$A < 0 + A \Rightarrow\Leftarrow$$.

This conclusion doesn't seem right. If you know

$$\lim_{n\to\infty} X_n = X,$$

$$\lim_{n\to\infty} Y_n = Y,$$

and

$$X_n < Y_n,\quad \forall\;n\in\mathbb{N},$$

then it only follows that

$$X \leq Y.$$

The inequality $$X < Y$$ is not necessarily true. So in your case, the limit gives you an inequality

$$A \leq 0 + A.$$

 

[jostpuur]

From the triangle inequality for measures we get

$$\mu(\left\{f>h\right\}) = \mu(\left\{f-f_{n}+f_{n}>h\right\}) \leq\mu(\left\{f-f_{n}>0\right\}) + \mu(\left\{f_{n}>h\right\})$$.

I don't think the "triangle inequality" was right either.

$$\Big(\;f(x)-f_n(x) > 0\quad\textrm{and}\quad f_n(x)>h\;\Big)\quad\implies\quad f(x)-f_n(x) + f_n(x) > h$$

so

$$\{x\in X\;|\; f(x)-f_n(x) > 0\}\;\cap\;\{x\in X\;|\; f_n(x)>h\} \;\subset\; \{x\in X\;|\; f(x)-f_n(x) + f_n(x) > h\}.$$

I don't see how you get that inequality from this. Of course it could be, that you did it some other way... I don't know about it at the moment. Perhaps you could explain in more detail how you thought that should work?

 

[jostpuur]

Here are key steps of my proof.

Assume that there is $$\delta > 0$$ so that

$$\mu(\{x\in X\;|\; f(x) > h\}) = A + \delta > A.$$

Prove that there exists $$\epsilon > 0$$ so that

$$\mu(\{x\in X\;|\; f(x) > h + \epsilon\}) > A + \frac{\delta}{2}.$$

To accomplish this, to my knowledge the only way is to assume that such epsilon would not exist, and then use the standard result that if there is a sequence of measurable sets

$$X_1\subset X_2\subset X_3\subset\cdots$$

then

$$\mu(\bigcup_{n=1}^{\infty} X_n) = \lim_{n\to\infty} \mu(X_n),$$

and arrive at contradiction.

Then prove the inclusion

$$\{x\in X\;|\; f(x) > h+\epsilon\} \subset \{x\in X\;|\; f_n(x) > h\}\;\cup\; \{x\in X\;|\; |f(x) - f_n(x)| \geq \epsilon\}.$$

This should be close to the goal.

 

[johnson123]

After proving the inclusion I get $$A + \delta/2 < A+\epsilon \Rightarrow \delta/2 < \epsilon$$

From this how can I arrive at a contradiction.

 

[jostpuur]

After proving the inclusion I get $$A + \delta/2 < A+\epsilon \Rightarrow \delta/2 < \epsilon$$

You don't get precisely that very naturally.

First: How far did you get in proving that the epsilon originally exists? I skipped all details of the proof in homework helping spirit, but it is a highly non-trivial proof.

 

[johnson123]

I agree it is not a simple fact. I did not thoroughly go through that result, I am more interested in seeing the idea(punchline) in the proof you kindly posted.

 

[jostpuur]

hmhmhm... I'm not sure what I should add to this anymore. It's not allowable to give complete solutions here... and it's so relative that when hints are fine and when there is too much of them... hmhmh... Considering your earlier mistakes with limits and triangle inequality, IMO you just must put more time into this. You know. Fight with the problem for hours and hours, keep breaks, then come back to the problem and so on, and see how things progress That's the only way to learn.

 

[johnson123]

triangle inequality for measures.

Theorem: $$\mu ( \left\{ \Sigma f_{n} > \Sigma \epsilon_{n}\right\} ) \leq \Sigma \mu(\left\{ f_{n} > \epsilon_{n} \right\})$$

Proof: If for all n we have $$f_{n} \leq \epsilon_{n}$$ then

$$\Sigma f_{n}(x) \leq \Sigma \epsilon_{n}$$ and so

$$\bigcap \left\{f_{n} \leq \epsilon_{n} \right\} \subseteq \left\{ \Sigma f_{n} \leq \Sigma \epsilon_{n} \right\}$$

This Implies $$\left\{ \Sigma f_{n} > \Sigma \epsilon_{n} \right\} = \left\{ \Sigma f_{n} \leq \Sigma \epsilon_{n} \right\}^{c} \subseteq ( \bigcap \left\{f_{n} \leq \epsilon_{n}\right\})^{c} = \bigcup \left\{f_{n} > \epsilon_{n} \right\}$$.

 

[jostpuur]

Oh, sorry. The triangle inequality was right then. I hadn't seen it before.