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Thread: Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second E

  1. #1
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    Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second E

    Hello all I was presented with this question from Folland's real analysis second edition on Radon measures which I am stuck on and so would really appreciate the help on.
    I m a novice in Radon measures especially in the concepts and abstractions so any help would be appreciated. It is problem #7 on page 220
    It reads as follows:
    Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second E-questionfolland.png
    Just in case, the definition of Radon Measure used in the book is:
    A Radon measure on X is a Borel measure that is finite on all
    compact sets, outer regular on all Borel sets, and inner regular on all open sets.
    I also know for a fact that Radon measures are also inner regular on all their sigma finite sets.
    I have tried to attack the problem many times but to no avail as for some reason I cannot seem to incorporate the given assumption that X is sigma finite.
    Also the assumption is X is locally compact Hausdorff space.
    I would really appreciate the help Thanks
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  2. #2
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    Re: Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Seco

    What I have done so far is: obviously the new measure is finite on all compact sets and also outer regular on alll Borel sets all is left for me to show is it is inner regular on all open sets
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  3. #3
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    Re: Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Seco

    Have $\displaystyle \mu$ a $\displaystyle \sigma \text{-finite}$ Radon measure on $\displaystyle X$.

    For each $\displaystyle A \in \mathscr{B}_{X}$, define a Borel measure $\displaystyle \mu_{A}$ by $\displaystyle \mu_{A}(E) = \mu(E \cap A)$.

    Show that, for each $\displaystyle A \in \mathscr{B}_{X}$, $\displaystyle \mu_{A}$ is in fact a Radon measure.

    Given Definition: A Radon measure on X is
    1. A Borel measure.
    2. Finite on all compact sets.
    3. Outer regular on all Borel sets.
    4. Inner regular on all open sets.

    (Note that you def differs from the one I looked up on wikipedia https://en.wikipedia.org/wiki/Radon_measure). Yours looks like a "moderated Radon measure".)

    I haven't done this stuff in a while, so I of course might have bungled something, and there might be a much better way to do it. But this is how I'd do it:

    1.) Everything in sight will be either compact (in a Hausdorff space, hence closed), open, or Borel - or some complement, or some at most countable intersection or union of such. So everything in sight is Borel, and so I won't bother mentioning it again. (All the definitions I encountered assumed the ambient space is Hausdorff).

    2.) The way I've done it has nothing to do with the specific properties of open sets or compact sets - but instead everything to do with $\displaystyle \mu_A$ eventually inheriting its properties from $\displaystyle \mu$, properties like $\displaystyle \mu$ being inner regular on open sets.

    That's just an observation - the kind of thing whose utility comes in checking the work (and sometimes streamlining it). If you had strong reason to suspect, for instance, that the specific compactness properties were crucial to proving that $\displaystyle \mu_A$ is inner regular on open sets, then you'd have strong reason to suspect that I'd made a mistake somewhere - or, if I made no mistake, then you'd gain a new insight into why compactness itself actually wasn't crucial here as you had previously thought.

    3.) Motivating Case:

    Rather than just throw a mess at you, I'll explain how I came to that mess. (I always these measure theory manipulations & techniques a bit confusing - sometimes applying a "geometric" insight only after I'd actually found the proof via technique/manipulation rather than motivation. So I appreciate that it can be seem like a messy jumble to read these things.)

    Suppose $\displaystyle U$ is open such that $\displaystyle \mu(U) < \infty$. Then since $\displaystyle \mu$ is inner regular on open sets:

    Given any $\displaystyle \epsilon > 0$, there exists a compact $\displaystyle K \subset U$ such that $\displaystyle \mu(K) \le \mu(U) < \mu(K) + \epsilon$.

    Also, everything set in sight in what follows will be a subset of $\displaystyle U$, and so have finite $\displaystyle \mu$ measure. (The $\displaystyle A$ and $\displaystyle A^c$ will be intersected resulting in a subset of $\displaystyle U$).

    Now $\displaystyle U = (U \cap A) \cup (U \cap A^c)$ and $\displaystyle K = (K \cap A) \cup (K \cap A^c)$, and both are disjoint unions, thus

    $\displaystyle \mu(U) = \mu(U \cap A) + \mu(U \cap A^c)$ and $\displaystyle \mu(K) = \mu(K \cap A) + \mu(K \cap A^c)$, so

    $\displaystyle \[ \mu(U) - \mu(K) \] = \[ \mu(U \cap A) - \mu(K \cap A) \] + \[ \mu(U \cap A^c) - \mu( K \cap A^c) \]$.

    Notice that with each of those braces, you have the measure of a set minus the measure of a subset of it - and that larger set of the (finite measure) $\displaystyle U$

    Thus each of those braces will have a value that's finite and non-negative.

    Therefore $\displaystyle 0 \le \mu(U \cap A) - \mu(K \cap A) \le \mu(U) - \mu(K)$. But $\displaystyle \mu(U) - \mu(K) < \epsilon$.

    Combine that with the definition of $\displaystyle \mu_A$ and it reads: $\displaystyle 0 \le \mu_A(U) - \mu_A(K) < \epsilon$.

    So, given any $\displaystyle \epsilon > 0$, I've produced a compact $\displaystyle K \subset U$ such that $\displaystyle \mu_A(K) \le \mu_A(U) < \mu_A(K) + \epsilon$.

    Thus $\displaystyle \mu_A(U) = \sup \{ \mu_A(K) \ | \ K \subset U, K \text{ compact }\}$.

    4.) The issue is how to handle when $\displaystyle \mu(U) = \infty$.

    If you're a mathematician trying to dream up a theorem out of the blue, you'd say to yourself "I have it in the finite case... so I'm going to need something, some condition or property, that might allow me to use my finite-case result to transition to the arbitrary case where it might be infinite." There are a lot of conditions and approaches to doing that - compactness (and its kin: locally commpact, paracompact, yadda yadda) comes immediately to mind. You already have a compactness property, from the Radon measure being inner regular - so maybe you don't need anything additional, and that suffices. Or maybe not, and you need an additional assumption (this kind of "guess the theorem" activity is why mathematicians need to have a huge list of examples in their heads in order to quickly rule out possibilities). There are other types of properties that might allow you to go from true-for-finite-cases to true-for-all-cases - like maybe requiring that the space is first or second countable. Since you're dealing with measures, requiring that the measure is finite does it instantly. The wikipedia def requires that Radon measures are locally finite... maybe that would do it.

    As a math student trying to find a proof of a given theorem, things are much easier. You have a big hint here, because theorems aren't presented with extraneous conditions, so, as a rule of thumb, at some point in your work, you'll need to use every single condition given. There can be exceptions to that of course, but as a general principle, it's really helpful to keep in mind.

    Both of these considerations, from a mathematician's guess-the-theroem need to somehow extend true-when-finite to true-even-when-infinite, and the math student's maxim "I'll need to use every condition" scream that, at this point, you're going to need to use that $\displaystyle \mu$ is $\displaystyle \sigma \text{-finite}$.

    5.) Since $\displaystyle \mu$ is $\displaystyle \sigma \text{-finite}$, there exists a countable set of measureable (here, Borel) sets of finite measure covering the entire space.

    There exists $\displaystyle \{E_k\}_{k = 1}^{\infty} \subset \mathscr{B}_{X}$ such that $\displaystyle X = \cup_{k = 1}^{\infty} E_k$ and $\displaystyle \mu(E_k) < \infty \ \ \forall \ k$.

    Now use that your def of a Radon measure says that it's outer-regular on all Borel sets to find an open cover $\displaystyle \{V_k^*\}_{k = 1}^{\infty} \subset \tau_{X}$

    such that $\displaystyle X = \cup_{k = 1}^{\infty} V_k^*$ and $\displaystyle \mu(V_k^*) < \infty \ \ \forall \ k$.

    That's possible because $\displaystyle \mu(E_k) < \infty$ and $\displaystyle \mu(E_k) = \inf\{ \mu(\mathscr{O}) \ | \ E_k \subset \mathscr{O} \in \tau_{X} \}$,

    so can always find an open set $\displaystyle V_k^*$ containing $\displaystyle E_k$ such that $\displaystyle \mu(V_k^*) < \mu(E_k) + 1 < \infty$.

    Now do the standard trick where you make that cover increasing by defining $\displaystyle V_n = \cup_{k = 1}^{n} V_k^* \ \ \forall \ n \ge 1$, so that $\displaystyle V_n \subset V_{n+1} \ \ \forall \ n \ge 1$.

    Since each $\displaystyle V_n$ is a union of a finite number of open sets, each of finite measure, if follows that it's also an open set of finite measure.

    So from the measure being outer-regular and $\displaystyle \sigma \text{-finite}$, it follows that:

    There exists $\displaystyle \{V_k\}_{k = 1}^{\infty} \subset \tau_{X}$ such that $\displaystyle X = \cup_{k = 1}^{\infty} V_k$, and for all $\displaystyle n \ge 1, \ V_n \subset V_{n+1}$ and $\displaystyle \mu(V_n) < \infty$.

    (Note that that cover could be finite. In the minimal case, if $\displaystyle \mu(X) < \infty$, it could be $\displaystyle X = V_1 = V_2 = V_3 = \dots$.)

    6.) Now let $\displaystyle U$ open. Using the $\displaystyle \sigma \text{-finite}$ open cover $\displaystyle \{V_k\}_{k = 1}^{\infty}$ of $\displaystyle X$ from #5, define $\displaystyle U_n = U \cap V_n$.

    Then for all $\displaystyle n \ge 1$, have that $\displaystyle U_n$ is open and $\displaystyle \mu(U_n) < \infty$. Also, from $\displaystyle V_n \nearrow X$, get that $\displaystyle U_n \nearrow U$.

    From $\displaystyle U_n \nearrow U$, get that $\displaystyle (A \cap U_n) \nearrow (A \cap U)$, and so $\displaystyle \mu_A(U_n) \nearrow \mu_A(U)$.

    Since $\displaystyle U_n$ is open and has finite measure, the motivating case #3 proves that $\displaystyle \mu_A(U_n) = \sup \{ \mu_A(K) \ | \ K \subset U_n, K \text{ compact }\}$.

    Thus $\displaystyle \forall \ \epsilon > 0, n \ge 1, \exists \ K_{n, \epsilon} \subset U_n \subset U$ such that $\displaystyle K_{n, \epsilon}$ is compact, and $\displaystyle \mu_A(K_{n, \epsilon}) \le \mu_A(U_n) < \mu_A(K_{n, \epsilon}) + \epsilon$.

    (Combining that with the fact that $\displaystyle \mu_A(U_n) \nearrow \mu_A(U)$ means that, whether $\displaystyle \mu_A(U_n)$ is going to a finite value or infinity, we can always find compact subsets close to it, so whose measures are going along along for the same ride.)

    Case I: $\displaystyle \mu_A(U) = \infty$.

    Fix $\displaystyle M >0$.

    Then, since $\displaystyle \mu_A(U_n) \nearrow \mu_A(U) = \infty$, there exists $\displaystyle N$ such that $\displaystyle n \ge N$ implies that $\displaystyle \mu_A(U_n) \ge M + 1$.

    Since for all $\displaystyle n \ge 1$, have that $\displaystyle \mu_A(K_{n, 1}) \le \mu_A(U_n) < \mu_A(K_{n, 1}) + 1$, it's also always true that $\displaystyle \mu_A(K_{n, 1}) > \mu_A(U_n) - 1$.

    Thus $\displaystyle n \ge N$ implies $\displaystyle \mu_A(K_{n, 1}) > \mu_A(U_n) - 1 \ge M$.

    Thus given any $\displaystyle M >0$, have produced a compact subset $\displaystyle K_{N, 1}$ of $\displaystyle U$ such that $\displaystyle \mu_A(K_{N, 1}) \ge M$.

    Thus $\displaystyle \sup \{ \mu_A(K) \ | \ K \subset U, K \text{ compact }\} = \infty$, and so $\displaystyle \mu_A(U) = \sup \{ \mu_A(K) \ | \ K \subset U_n, K \text{ compact }\}$.

    Case II: $\displaystyle \mu_A(U) < \infty$.

    Fix $\displaystyle \epsilon > 0$.

    Then, since $\displaystyle \mu_A(U_n) \nearrow \mu_A(U) < \infty$, there exists $\displaystyle N$ such that $\displaystyle n \ge N$ implies that $\displaystyle 0 \le \mu_A(U) - \mu_A(U_n) < \frac{\epsilon}{2}$, so that $\displaystyle \mu_A(U_n) \le \mu_A(U) < \mu_A(U_n) + \frac{\epsilon}{2}$.

    But for all $\displaystyle n \ge 1$, have that $\displaystyle \mu_A(K_{n, \epsilon /2}) \le \mu_A(U_n) < \mu_A(K_{n, \epsilon /2}) + \frac{\epsilon}{2}$.

    Thus $\displaystyle n \ge N$ implies that:

    $\displaystyle \mu_A(K_{n, \epsilon /2}) \le \mu_A(U_n) \le \mu_A(U) < \mu_A(U_n) + \frac{\epsilon}{2} < \left( \mu_A(K_{n, \epsilon /2}) + \frac{\epsilon}{2} \right) + \frac{\epsilon}{2}$ $\displaystyle = \mu_A(K_{n, \epsilon /2}) + \epsilon$.

    Which says that, if $\displaystyle n \ge N$, then $\displaystyle \mu_A(K_{n, \epsilon /2}) \le \mu_A(U) < \mu_A(K_{n, \epsilon /2}) + \epsilon$.

    Thus, for all $\displaystyle \epsilon >0$, there exists a compact $\displaystyle K \subset U$ such that $\displaystyle \mu_A(U) < \mu_A(K) + \epsilon$, and so, since $\displaystyle U$ has finite measure greater than all of its compact subsets,

    if follows that $\displaystyle \mu_A(U) = \sup \{ \mu_A(K) \ | \ K \subset U_n, K \text{ compact }\}$.

    QED
    Last edited by johnsomeone; Jul 27th 2015 at 04:03 PM.
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  4. #4
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    Re: Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Seco

    @johnsomeone : Thank you very much mate finally got it
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