# Tag Info

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The question leaves your background a bit unclear, so I chat descriptively about the three Haar measures you are likely to be aware of without even knowing that they are Haar measures. Nothing in what follows is rigorous, but rather seeks to give you a taste of what Haar measure is about. We can measure the size $m(S)$ of a subset $S$ of $\mathbb{R}$ simply ...

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People use measure theory in tandem with differential forms all the time—there's no contradiction whatsoever between the formalisms. Be aware, though, that the adjective “Riemannian” in the context of differential geometry refers to constructions depending on Riemannian metrics (which are “Riemannian” in the sense of originating in the work of Bernhard ...

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Let throughout this post $(\Omega,\mathcal{F},P)$ be a probability space, and let us first define the conditional expectation ${\rm E}[X\mid\mathcal{G}]$ for integrable random variables $X:\Omega\to\mathbb{R}$, i.e. $X\in L^1(P)$, and sub-sigma-algebras $\mathcal{G}\subseteq\mathcal{F}$. Definition: The conditional expectation ${\rm E}[X\mid\mathcal{G}]$ ...

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Counting measure is just summation! To see this, you can approach from a few different angles; how about we consider the Monotone Convergence Theorem. To that end, for $n\in\mathbb{N}$, define $f_n:\mathbb{N}\to\mathbb{R}$ by $$f_n(k)=\begin{cases}f(k) & \text{if }1\leq k\leq n\\ 0 & \text{else}\end{cases}.$$ Then clearly, as $n\to\infty$, $f_n\... 62 For$\epsilon > 0$,$\mu((q_n - \epsilon, q_n + \epsilon)) = 2\epsilon$and$\sum_{n=1}^\infty 2\epsilon = \infty$, so this reasoning doesn't work. A simple method to show that$\mu(\mathbb Q) = 0$is to notice that the measure of a single point is$0$, thus: $$\mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\})... 43 A probability on \mathbb{R}, be it continuous or not, is given by its CDF x \mapsto\mathbb{P}(X \leq x). A CDF is right-continuous, and the set of right-continuous functions has the cardinality of \mathbb{R}. To see this, you can for instance argue that the values of such a function are given by its values at the rational points, so it has at most the ... 41 Nothing bad happens if you allow a metric to assume the value \infty . In fact, only good things happen. Nothing bad happens since you can still talk about open balls, the induced topology, uniformly continuous functions, continuous functions, Lipschitz functions etc. and the theory looks pretty much the same as it would if the metric is not allowed to ... 41 Hint: If f is monotone, then, for every real number x, the set$$f^{-1}((-\infty,x])=\{t\mid f(t)\leqslant x\}$$is either \varnothing or (-\infty,+\infty) or (-\infty,z) or (-\infty,z] or (z,+\infty) or [z,+\infty) for some real number z. To show this, assume for example that f is nondecreasing and that u is in f^{-1}((-\infty,x]), ... 37 "As the rationals are dense in \mathbb{R} we must have \mathbb{R}\subset \bigcup_{n=1}^\infty I_n." This is false. Pick your favorite irrational number x. For every n, there exist infinitely many rationals q_m such that q_m is within a_n/2 of x. But the least such m may be much larger than n. In particular, it does not follow that there ... 36 Because B_x is supposed to be the intersection of all measurable sets containing x, but you've found a measurable set containing x strictly inside B_x. Because for any measurable set T, we have T=\bigcup_{x\in T}B_x. Thus, if there are n distinct sets of the form B_x, then there are at most 2^n elements of S. 36 Ok so there is a lot to say here, let's start with the easiest question. The use of C^+_K instead of C_K change absolutely nothing because we can always write f=f^+-f^- with f^+=\max(f,0) and f^-=\max(-f,0) which are positive and integrable. In order to avoid confusions between the different names of the different convergence i will talk about ... 35 You're imagining that the open intervals of G^c are ordered discretely, like the integers, so you have alternating open intervals in G^c and closed intervals in G. But actually, the open intervals of G^c are densely ordered, and order-isomorphic to the rationals. As a result, there is no "next (...) in line for each [...]" like you claim there is. ... 35 Yes. Here is a proof over \mathbb{C} . Matrices with repeated eigenvalues are cut out as the zero locus of the discriminant of the characteristic polynomial, thus are algebraic sets. Some matrices have unique eigenvalues, so this algebraic set is proper. Proper closed algebraic sets have measure 0. (intuitively, a proper closed algebraic set is a ... 34 Suppose x_0\in A. Let B := \{ d(x,K) < 1\}. Observe that since K is compact, there exists y_0\in K such that d(x_0,y_0) = 1. By definition B_1(y_0) = \{x: d(x,y_0) < 1\} is a subset of B. This implies that for all \epsilon < 1/2, we have that$$ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $$where \mu is ... 34 Given a measure space (X_1,M_1,\mu) and a measureable space (X_2,M_2) you can define the pushforward measure on M_2 of \mu by a measurable function F:X_1\to X_2 to be F\mu(E)=\mu(F^{-1}(E)). Then you have the formula$$\int_{X_2}g\;\mathrm{d}F\mu=\int_{X_1}g\circ F\;\mathrm{d}\mu$$which is effectively the change of variables between the ... 34 You could utilize one of the well known ways to count the rational numbers, namely consider the integer lattice \mathbb Z^2 and the subset \{(a,b)\mid a\geq 1\ \wedge\ b\geq 0\} as illustrated here: This corresponds to the positive rationals, namely (a,b)\mapsto\frac ba. It is a surjective covering and it is now simple to see how we might cover all ... 33 Start with the definition. The Lebesgue integral of a simple function s = \sum_{j=1}^n \alpha_j \ \chi_{A_j} is:$$ \int_E s \,d\mu = \sum_{j=1}^n \alpha_i \ \mu(E \cap A_j) $$If \mu(E) = 0, then \mu(E \cap A_j) = 0 for all j. Thus \int_E s \,d\mu = 0. The Lebesgue integral of a nonnegative function f is the supremum of integrals of all ... 33 Let f:X\to \mathbb{R}. Assume that f is measurable, and that \|f\|_p<\infty for all large p. Suppose for convenience that f\geq 0. (If not, just work with f^*:=|f|.) We define$$ \|f\|_{\infty}:=\sup \{r\in \mathbb{R}: \mu\left( \{x:|f(x)|\geq r\} \right)>0\}. $$I claim without proof that \|f\|_p < \infty for large p implies that ... 33 This is a really hard question; I think in general intuition for this sort of thing tends to come with experience, as you get used to the concepts. Having said that, I'll try to articulate the way that I think about it. I guess the way of viewing \mathbb{Q} as a subset of \mathbb{R} is a load of dots on a continuous line. Obviously these dots are very ... 32 Let (X,\mathcal{A},\mu) be a measure space and (f_n)_{n \in \mathbb{N}} such that f_n \to f in measure, i.e.$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$for any \varepsilon >0. Setting \varepsilon=2^{-k}, k \in \mathbb{N}, we can choose n_k such that$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$for all n \geq n_k. Without ... 32 Let A, B be two closed subsets of [0, 1], both with measure \frac{1}{2}, and suppose A\cap B = \emptyset. Then$$m([0, 1]\setminus(A\cup B)) = m([0, 1]) - (m(A) + m(B)) = 1 - \left(\tfrac{1}{2} + \tfrac{1}{2}\right) = 0.$$But [0, 1]\setminus(A\cup B) is open and the only open set with measure zero is the empty set, so [0, 1]\setminus(A\cup B)... 32 Another simple example. The natural filtration when we model tossing a die twice in a row. Here is how it works. Let X_1 be the outcome of the first toss. So the values of X_1 are in the set \{1,2,3,4,5,6\}. Let X_2 be the outcome of the second toss. As a sample space, we can take \Omega = \{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}, the set of ... 30 The standard example is given by the function g(x)=f(x)+x, where f is the devil's staircase function of Cantor. It turns out that the function g is a homeomorphism from [0,1] onto [0,2] and has the property that \mu(g(C))=1 (where C is the Cantor set). Pick a non measurable A\subset g(C). First note that B=g^{-1}(A) is measurable since ... 30 There is a easy way to show that. Suppose that p<q and X a space measure finite. Take any f\in L^q. Then, the q-norm is finite. In this way,$$\int_X|f|^p = \int_{f(x)<1}|f|^p + \int_{f\geq1}|f|^p \leq\int1+\int_{f\geq1}|f|^q\leq\mu(X)+||f||_q^q<\infty$$A counter example just take$$f(x)=\frac{1}{x}$$for x\in(0,\infty) and Lebesgue measure. ... 29 Let (r_n)_{n\in\mathbb{N}} be a dense sequence in \mathbb{R} (it could, for example, be an enumeration of the rationals). For k \in \mathbb{N}, let$$U_k = \bigcup_{n\in\mathbb{N}} (r_n - 2^{-(n+k)},\, r_n + 2^{-(n+k)}).$$U_k is a dense open set with Lebesgue measure \leqslant 2^{2-k}, thus$$N = \bigcap_{k\in\mathbb{N}} U_k$$is a set of the ... 29 There is no universal and objective definition of what is a measurable subset of a general space$X$. The general concept of a measurable subset has its origins in the problem of measure in Euclidean space: Problem of measure: Given an object$A\subset\mathbb R^n$, how does one assign a measure$m(A)\in[0,\infty]$to$A$? (In the case$n=1,2$and$3$, ... 29 Probability when there are only finitely many outcomes is a matter of counting. There are$36$possible results from a roll of two dice and$6$of them sum to$7$so the probability of a sum of$7$is$6/36$. You've measured the size of the set of outcomes that you are interested in. It's harder to make rigorous sense of things when the set of possible ... 28 Let$(\Omega,\mathcal{F},P)$be a probability space, i.e.$\Omega$is a non-empty set,$\mathcal{F}$is a sigma-algebra of subsets of$\Omega$and$P:\mathcal{F}\to [0,1]$is a probability measure on$\mathcal{F}$. Now, suppose we have a function$X:\Omega\to\mathbb{R}$and we want to "measure" the probability of$X$belonging to some subset of$\mathbb{R}\$. ...

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