7

You can always get a finite measure on $X$ by choosing your favourite $p \in X$ and defining $\mu(E) = 1 \iff p \in E$ and $\mu(E)=0$ otherwise. Note this measure has nothing to do with the topology on $X$ since $\mu$ is defined for both Borel and non-Borel subsets. In case this measure is not the sort of thing you're looking for, you should put some ...


7

Without further restrictions, it's not true in general. For a silly example, consider the set $X=\{a, b\}$ with the discrete topology, and the Borel measure $\mu$ on $X$ given by $$\mu(\emptyset)=\mu(\{a\})=0,\quad \mu(\{b\})=\mu(X)=1.$$


6

Not even for Lebesgue measure on $\mathbb{R}$. Enumerate the rationals as $q_n$ and for each $\epsilon > 0$ let $U_\epsilon = \bigcup_{n=1}^\infty (q_n - \frac{\epsilon}{2^n}, q_n + \frac{\epsilon}{2^n})$. Clearly $U_\epsilon$ is open and dense, and its Lebesgue measure is at most $2\epsilon$. Now consider $G = \bigcap_{k=1}^\infty U_{1/k}$. Then $G$ ...


4

Let $A_n$ be defined as in the lemma. Fix $\epsilon>0$ and choose $K_n \in \mathfrak{R}$, $\overline{K_n}$ compact and $\overline{K_n} \subseteq A_n$, such that $\mu(A_n)<\mu(K_n) + \epsilon/2^n$. By what you've already shown, there's an $N$ such that $\bigcap_{n=1}^N K_n = \emptyset$. Now, \begin{align} \mu(A_N) &= \mu(\cap_{n=1}^N A_n) \leq \mu(\...


4

Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $\mathfrak c$ Borel subsets of $C$, but $2^{\mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is ...


3

The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $\mathcal{B}( \mathbb R)$, every countable subset of $\Bbb R$ is a countable union of singletons.


3

The Borel $\sigma$-algebra on any space is complete with respect to counting measure, since the only null set for counting measure is the empty set. However, the Borel $\sigma$-algebra on $\mathbb{R}$ is not complete with respect to any $\sigma$-finite measure. Note that $\mathbb{R}$ is Borel-isomorphic to $\mathbb{R}^2$ (see Are the measurable spaces $(\...


3

Your $F$ and $G$ are holomorphic in $\Omega=\{\Re z>0\}$ (a clean way to see this is to first show they're continuous and then apply Morera's Theorem). Alas $0$ is just a boundary point of $\Omega$, so the argument doesn't quite work, or at least it's not quite that simple. One way to fix it is to show that $F$ and $G$ are bounded and to note that the ...


3

The Borel-$\sigma$-Algebra contains any subset that can be formed by intersecting or taking countable unions of open sets. I assume you want your finite set to carry the discrete topology, i.e. every set is open. Then you're Borel-$\sigma$-Algebra will just be the entire power set.


2

Hint: for any $a$, you can write $\{x\in X: f(x)<a\}$ as a countable union of (measurable) sets of the form $f^{-1}([-\infty,r))$. Then use your hypothesis and the fact that a countable union of measurable sets is measurable to conclude.


2

Find a sequence $(r_{n})$ of rational numbers such that $r_{n}<a$, and $r_{n}\uparrow a$, then $f^{-1}([-\infty,a))=\displaystyle\bigcup_{n=1}^{\infty}f^{-1}([-\infty,r_{n}))$.


2

Your inequality is backwards. Just note that $E_n \subseteq \bigcup_{k=n}^{+\infty}E_k$ so $$ {\lim}_{n\to\infty}m\left(\bigcup_{k=n}^{+\infty}E_k\right) = \limsup_{n\to\infty} m\left(\bigcup_{k=n}^{+\infty}E_k\right) \ge \limsup_{n\to\infty} m(E_n)$$ Also have a look here.


2

I haven't looked at it in detail but the book by Fremlin has a section covering exactly what you want. You can find it on the author's website. I'll just sketch some of the ideas here and refer to Fremlin's book for the details. In section 441.B the author discusses a theorem of Steinlage which is a generalization of the existence of Haar measures: ...


2

Yes. Let $X$, $Y$ non-empty sets, $A$ a $ \sigma-$ algebra on $X$, $B$ a $ \sigma-$ algebra on $Y$ and $f:X \to Y$ a mapping. If $C$ is a collection of subsets of $B$ such that $\sigma(C)=B$, then we have: $f$ is $(A,B)-$ measurable $ \iff f^{-1}(M) \in A$ for all $M \in C$.


2

Finite measures always exist, on any set. Most of the time you at least want the Borel sets to be measurable to have any topology-measure connection. There is a sizable theory on topological measure theory (see Fremlin's book, part 4 of 5 volumes on measure theory in general..) about such subjects. There is a lot of interaction between measures of certain ...


2

It's hard to answer without further information. A few things come to mind: The Lebesgue measure is the unique translation invariant regular Radon measure $\mu$ (translation invariant means s.t. $\mu(A)=\mu(x+A)$ for $x\in\mathbb{R}^n$ and $A\subset\mathbb{R}^n$ measurable) on $\mathbb{R}^n$ such that $\mu([0,1]^n)=1$ (it is the Haar measure on $\mathbb{R}^...


2

$\{u\le a\}$ indeed denotes the preimage $u^{-1}((-\infty,a])=\{x:u(x) \le a\}$. This is going to be $(-\infty, \lfloor a\rfloor]$ which is an element of $\mathcal B(\Bbb R)$. For your second example, $v:\Bbb R^2\to\Bbb R$, so it's rather the single $(-\infty, a]$ rays that generate $\mathcal B(\Bbb R)$, and the '?' is to be filled by $a$. The preimage of $...


2

Let $\mu_X =\mu_Y=$ Lebesgue measure on $[0,1]$ and $ \mu = \mu_X \times \mu_Y$. Then $\mu \{(x,y):x=y\}=0$. Hence $\mu \{(x,y):|x-y| <\delta \} \to 0$ as $\delta \to 0$. Let $E$ be the complement of $\{(x,y):|x-y| <\delta \}$ and choose $\delta$ such that $\mu (E) >1-\epsilon$. Suppose there are sets $A$ , $B$ in $[0,1]$ such that $x \in A$ and $y ...


2

Try with the counting measure and the indicator function of the rationals.


2

Yes: in particular, it's a countable sum of step functions, so is a uniform limit of step functions, so is regulated, and all regulated functions are measurable.


2

$\sqrt{x}$ and $1/\sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.


2

Once we establish $b \Rightarrow c$, then the rests follow directly. Assume to the contrary that $|f|>0$ on a set of positive measure. Then there exists a ball $B\subset\mathbb{R}^n$ such that $$ \int_B |f(x)|dx=c >0. $$ Without loss of generality, assume that $B= B(0,1) =\{x\in\mathbb{R}^n\;|\;\|x\|<1\}$. Then, for $y$ with $\|y\|=r$, we have $$\...


2

I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of $\{\infty\}$, $\{-\infty\}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.


2

We have the standard Lebesgue measure $\mu$, that is defined on all Borel sets, and obeys translation invariance and interval-consistency (my name for $\mu((a,b])=b-a$ for all relevant intervals; you call it uniform in the comments) . Indeed there are $\mathfrak{c}$ many Borel sets on which this $\mu$ is then defined. Independently of that we can define ...


2

The Lebesgue measure is simply the closure of the Borel measure. As an analogy, think about the number line. Basic operations give us the rational numbers, and that is all fine and dandy as far as basic algebra is concerned. But once we start thinking about things such as topology, we add in extra numbers called the irrational numbers to "close" the gaps in ...


2

Suppose that $J_{n}(B)$ is infinite. Take the union over $J$: $\begin{equation}\mu(\bigcup\limits_{j \in J} B \cap A_{j}) > \sum_{j \in J} \frac{1}{n} = \infty \end{equation}$. Contradiction with the fact that $\mu$ is $\sigma$-finite.


2

Note that $f$ does not take on the values $0,-\infty.$ Note also that on $(-\infty,0)$, $f(x)=1/x$ is a bijection; same for $(0,\infty).$ Below we look at all cases of $a\in[-\infty,\infty].$ $a=-\infty:$ $f^{-1}([-\infty,a])=f^{-1}(\{-\infty\})=\emptyset,$ so we're fine. $-\infty<a<0:$ Here $f^{-1}([-\infty,a])=f^{-1}((-\infty,a]),$ which is the ...


2

It seems like a reasonable assumption. The converse can't be literally true. For a trivial counterexample, say $X$ is not second-countable but $\mu$ is supported on a second-countable subset of $X$.


2

See Theorem 4.13 in Brezis's functional analysis. If the measure space $(\Omega, \Sigma,\mu)$ is separable, meaning that the $\sigma$-algebra $\Sigma$ can be generated by a countable subset of $\Sigma$. then $L^p(\Omega)$ is separable (as a metric space) for $1\leq p<\infty$. Now suppose $\Omega=X$ is a second-countable topological space, $\Sigma=\...


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