Dears,  the question is - why there are finitely-additive measures in dual to L infinity, not only sigma-additive? I was thinking, that sigma-additive measures are related with embedded elements of L1 into (Linfinity)*, whereas finitely-additive [and not sigma-additive] are related with "the rest". Can someone explain that or give me link to article or pdf where it is well explained? I want to know what gives that difference.
120724
Can we identify the dual space of $l^\infty$ with another "natural space"?  If the answer is yes, what can we say about $L^\infty$? By the dual space I mean the space of all continuous linear functionals.
53741
Given a measure zero set $E$, by definition we have forall $\varepsilon &gt; 0$, a covering of $E$ by intervals whose lengths sum to $&lt; \varepsilon$.  I want to prove that we can cover $E$ in intervals such that the sum of the lengths the intervals is $&lt; 1$ and each point in $E$ is contained in infinitely many of the intervals.  Do you know how to prove this? Thank you
87046
Let X be an uncountable set. And let $$\mathcal{A} =\{A\subset X: A \text{ is countable or } X-A \text{ is countable}\}.$$ Then it can be shown that $\mathcal{A}$ is a $\sigma$-algebra. Define $\mu:\mathcal{A}\rightarrow [0,\infty]$ by $\mu(A)=0$ if $A$ is countable or 1 otherwise. The task is to show that $\mu$ so defined is a measure on $\mathcal{A}$.  It is clear that $\mu(\emptyset)=0$. Now we want to show that $\mu$ is countably additive on disjoint sets from $\mathcal{A}$. I'm first trying to show this for two disjoint sets $A$ and $B$ in $\mathcal{A}$. I am running into a problem if I assume that both $X-A$ and $X-B$ is countable.  For if this happens then, $B\subset X-A$ implies $B$ is countable. But both $B$ and $X-B$ cannot be countable, since then $X=B\cup (X-B)$ will be countable, running contrary to the assumption that $X$ is uncountable. How should I address this case?
121971
I was wondering if someone could please give me an example of a sequence of decreasing sets where the first set has infinite Lebesgue measure; i.e., $\{B_{n}\}_{n=1}^{\infty}$ such that $m(B_{1}) = \infty$ but $m(\cap_{n=1}^{\infty} B_{n}) \neq \lim_{n \to \infty}m(B_{n})$?  thank you.
63270
Just having trouble with this problem. First, it says to prove that if $E$ is Lebesgue Measurable, and $\epsilon&gt;0$ is arbitrary, then there is an open $O$ such that $E \subset O$ and $m(O\setminus E)&lt;\epsilon$. Now for this part, since $E$ is Lebesgue measurable, I was able to take the definition of being Lebesgue outer measurable (the infimum of open coverings of $E$) and easily construct $O$ from that, and it works out.  But now I want to show that there is an $F$ closed with $F \subset E$ and $M(E\setminus F)&lt;\epsilon$. The problem is, I can't figure out how to construct this $F$! The idea seems obvious intuitively, but I feel like I am given no tools for constructing a closed subset of an arbitrary set that satisfies these conditions. Am I missing something obvious here?  Thanks!!
3742
The Sum of 2 sets of Measure zero might well be very large for example, the sum of $x$-axis and $y$-axis, is nothing but the whole plane. Similarly one can ask this question about Cantor sets:  If $C$ is the cantor set, then what is the measure of $C+C$?
56240
I have been struggling with the following for quite some time now. If anyone can give me some help, it will be much appreciated:  Let $f$ bounded, measurable and $E$ be a set of finite measure. Let $A \subset E$ be measurable. Prove that:  $\displaystyle \int_{A} f = \int_{E} f \chi_A $
24156
In the Lebesgue outer measure we have the following properties: the length of the empty set is $0$, monotonicity and countable subadditivity.  My question is, is it possible to use the above properties to show finite additivity?  I have no idea where to begin (or even if it's true) so any hints are definitely appreciated.
77245
I am a graduate student at Iowa State University attempting to return after a five-year hiatus and take the Real/Complex Analysis qualifier on January 8 for potential reinstatement. Since the professors here are on hiatus, I hope you guys don't mind if I bombard the board with a few past questions from our quals here, like this one from Spring 2013:      Let $m$ denote Lebesgue measure on $\mathbb R$ and $M$ the $\sigma$-algebra of Lebesgue measurable subsets. Given $E \in M$ with $m(E) &gt; 0$ and $\alpha \in (0, 1)$, prove that there is an interval $I$ such that $m(E \cap I) &gt; \alpha \, m(I)$.   I've attempted to force the issue using the definition of outer measure but it doesn't seem to work. I've also attempted using the definition of Lebesgue measurability but that doesn't seem to work either - so I'm at a loss!   Thanks in advance!   -Darrin Rasberry
77932
I am trying to solve following exercise from Folland,     If $\mu$ is a semifinite measure and $\mu(E) = \infty$, for any $C &gt; 0$, $\exists$ $F \subset E$ with $C &lt; \mu(F) &lt; \infty$.   It seems to follow from definition of semifinite measures, which you can find , but I couldn't prove it.
110983
Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?  I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true?  If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?  Any comments, references or tips would be greatly appreciated.
101000
Let $X\subset R$ and $Y \subset R$ where X and Y have finite cardinalities. Let also, $a,b \in R$. How to show that $|aX+bY|=|X||Y|$ almost everywhere (measure of $(a,b) \subset R^2$ such that $|aX+bY|&lt;|X||Y|$ is 0) ?  Where we define $|aX+bY|=\{ax+by:x \in X, y \in Y \}$.  More, specifically let $X=\{x=d_x*z: z \in \mathcal{Z}, d_x \in \mathcal{R}\}$ (so, it's a lattice. Where each point is $d_x$ away from one another.)  The above is easy if $X=Y$ then the proof of $|aX+bX|=|X||X|$ almost everywhere is fairly simple.   Now what if $Y$ is lattice with $d_y \neq d_x$. How to show it then? Thanks in advance.
120937
Prove, or give a counter example: Let $\mu$ be a finite positive borel measure on $\mathbb{R}$. Then $\int (x-y)^{-2} d \mu (y) = \infty $ almost everywhere on $\mu$ (for the selection of x's).  This is a question I had in an exam, and the answer is supposed to be presented in less than 30 words, so there must be something quite simple I'm missing.
30765
Let $P$ be a probability measure on a Borel $\sigma$-algebra (on some metric space, $\Omega$).  It is called tight if for every $\epsilon &gt;0$, there exists a compact $K$ such that $P(X \in K) \geq 1 - \epsilon$.   It is called inner regular if $P(A) = \sup \{P(K) | K \subset A, K \text{ compact}\}$ for every $A$ in the Borel $\sigma$-algebra.   It is clear that inner regularity implies tightness if we choose $A=\Omega$. For an arbitrary set $A$, using tightness we can obtain a compact $K(\epsilon)$ such that $P(A\cap K(\epsilon)) \geq P(A) - \epsilon$. If $A$ is closed then $K \cap A$ is compact and hence we get inner regularity for closed sets $A$. How can I prove this for general $A$?  Thanks.
60475
Monotone Convergence Theorem for general measure: Let $(X,\Sigma,\mu)$ be a measure space. Let $f_1, f_2, ...$ be a pointwise non-decreasing sequence of $[0, \infty]$-valued $\Sigma-$measurable functions, i.e. for every $k\ge 1$ and every $x$ in $X$, $$0 \le f_k(x) \le f_{k+1}(x).$$ Next, set the pointwise limit of the sequence ${f_n}$ to be $f$. That is, for every $x$ in $X$, $$f(x) = \lim_{k\to \infty}f_k(x).$$ Then $f$ is $\Sigma-$measurable and $$\lim_{k\to \infty}\int f_k d\mu = \int f d\mu.$$  I've noticed that when it comes to monotone convergence theorem (either Lebesgue or general measure), usually its definition restricts the monotone function sequences to be nonnegative. I'm not sure why the 'nonnegative' is necessary.
2479
Im doing a course on measure theory and I'm stuck on one of the exercises.  Take $\{Y_{\gamma}:\gamma \in C\}$ as an arbitrary collection of random variables and $\{X_{n}: n \in N\}$ to be a countable collection of random variables  I now want to show that $$\sigma \{Y_{\gamma} : \gamma \in C\}=\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$$  So i need to show that both $\sigma \{Y_{\gamma} : \gamma \in C\}\subset\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$ and $\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}\subset\sigma \{Y_{\gamma} : \gamma \in C\}$ hold.  I know that for a single random variable X you have $\sigma(X)=\{X^{-1}(B): B \in Borel\}$. How do I expand this for the collection of random variables?  I then also have to show that if $Fn=\sigma\{X_{1},...,X_{n}\}$ and $A=\cup_{n=1}^{\infty}F_{n}$ then $$\sigma(A)=\sigma\{X_{n}: n \in N\}$$  Can anyone help me with this?
38465
While reading the motivation of complete measure space , what I concluded was,  completeness is not really necessary when we define on one measure space  and it is necessary when we want to measure on product of measure spaces (is it true ?).  If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ?  I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ?  Essentially what is the flaw in the measure without being complete ?  Waiting for response. Thanks!