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You say the range of $f$ is contained in $[0,100]$ "for example". The weasel words make it hard to know the exact hypotheses. I'm going to assume $f\ge0$. One thing you could do is this: Say $$\frac1{\mu(A)}\int_A f=L.$$ Adding the fractions shows that $$\left|\frac1{1+f}-\frac1{1+L}\right| \le\frac{|f-L|}{1+L}.$$Hence ...

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Just look at the sequence $f_n(x) := \frac{1}{n}$ on the space $\Omega := [0,1]$ (endowed with Lebesgue measure). Since $f_n$ converges pointwise (even uniformly) to $f := 0$, we have in particular $f_n \to 0$ in measure. However, $$|\{x \in \Omega; |f_n(x)-f(x)|>0\}|=1,$$ so we cannot that this term converges to $0$ as $n \to \infty$.

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As already said, if $f_n\ge0$ this is just Fatou's Lemma. And in general the answer is no. But only because you asked the wrong question! Of course the answer is no if you're not assuming $f_n\ge0$; saying that the integrals are bounded says nothing about the size of $f_n$, the integral could be small because of cancellation. For that matter if the ...

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Using Fatou's lemma, if the functions $f_n$ are positive you don't need to suppose they are monotonic: $\int_X fdm \leq \liminf_n \int_Xf_n \leq \sup_n \int_X f_n dm < \infty$ If you don't suppose the functions positive, then the limit function may not be integrable. Take, for instance, the measure space $(X,\mathcal{B},m)=([-1,1], \mathcal{B}_{[-1,1]}, ... 1 Let's fix some notations. Denote the Haar measure on$\Bbb{T}^d$by$\lambda$instead of$m$. The measures suggested should probably be normalized by $$d\mu_{v,N}(y)=\frac{1}{(2N+1)^d}\left\|\sum_{m\in\mathbb{Z}^d, \lvert m_i \rvert \leq N}\chi_{-m}(y) U(m)v\right\|^2 d\lambda(y).$$ The characters of$\Bbb{T}^d$are the functions $$\chi_m: \Bbb{T}^d \to ... 0 When you're trying to prove (\Rightarrow), the limit gives you a way to bound the integrals of f_n by \epsilon for sufficiently large n. Then it's a matter of using the fact that any collection of finite number of L^1(E) functions are both uniformly integrable and tight. This is problem 1 in \textit{Royden and Fitzpatrick} page 99. In the ... 2 Michael Hardy has already explained why one wants to say 0\cdot(+\infty) is undefined (or indeterminate) in some contexts and 0 in at least one other context, namely integration. I think the value 0 is appropriate in geometric contexts too. Imagine a "rectangle" of 0 height but infinite width; it's essentially just a line, and its area, which ought to be ... 3 If f(x)\to 0 and g(x)\to\infty as x\to\text{something}, then f(x)g(x) could approach any number or +\infty or -\infty, depending on what functions f and g are. For example, consider \vphantom{\dfrac\int\int}x\cdot \dfrac 6 x as x\to 0. There is at least one context in which it makes sense to say 0\cdot\infty=0, and that is the theory ... 0 Each A_n is a finite union of some number, say q, of half-open intervals of the form (a,b]. Shrink each of these just a little from the left end; I'll decide later how much to shrink it but for now I'll just call the shrinkage amount \delta. So you're now looking at shrunk intervals (a+\delta,b] in place of (a,b]. Do this for each of the q ... 2 Of course the existence of the product measure is a standard thing in probability, following from the Kolmogorov Extension Theorem. Here you could cheat: \Omega is compact; you could define the product measure as a weak* limit of finite product measures, more or less. If I'm not missing anything the bit about the symmetric differences is immediate from a ... 1 In your definition of "doubling measure" you don't specify the centers of the balls. If you only assume the doubling condition for balls centered at the origin then an annulus can have measure zero and the bound you want doesn't exist. But of course typically one assumes that condition for balls with arbitrary centers. And now the point is that the reverse ... 1 Tempered distributions have Fourier transforms. If P(D)\,F=0, take the Fourier transform to get P(\xi)\,\hat F(\xi)=0 for all \xi\in\mathbb{R}. Since P(\xi)\ne0 if \xi\ne0 the support of \hat F is \{0\}. Can you take it fromhere? 6 Let \omega\colon\mathbb{R}\to[0,\infty) be measurable. Then$$ \mu_\omega(A)=\int_A\omega(x)\,dx $$defines a measure, which in general will not be translation invariant. If moreover \omega is periodic if period 1 and \int_0^1\omega(x)\,dx=1, then \omega will satisfy your requirements. 3 If it needn't be finite, then \mu(k)=1 for k\in\mathbb{Z} (and zero on \mathbb{R}\setminus\mathbb{Z}) has got this property. For example \mu(\{1\})\neq\mu(\{3/2\}). 1 You may prove by induction that \delta(\mathcal{G}_1),\dots,\delta(\mathcal{G}_k),\mathcal{G}_{k+1},\dots,\mathcal{G}_n are independent (for all k\le n)... For k=0 this is the assumption (\mathcal{G}_{1},\dots,\mathcal{G}_n are independent). For k>1 assume that ... 1 Michael gave a great answer, and I know this question was asked 3 years ago. However, I spent a long time trying to understand this, and for anybody else confused, I thought I would elaborate a little on Michael's answer: The main reference is these notes by Evans. For other helpful references, see Kozdron's notes, Zitkovic's notes, Gallager's notes, and ... 0 Disclaimer: I'm not quoting historical facts below. The issue, as I see it, is this: if one postulates the existence of a translation invariant measure on \mathbb R, one can show that such a measure cannot exist. Now, the intuitive notion of length in \mathbb R can be used to define something akin to a measure, by approximating the size of a set with ... 1 There are detailed answers to this question through the cases where Riemann integral fails, but I could never understand those clearly. I could try to answer your question this way: Lebesgue integral (of a complex-valued function) is defined through the integral of a positive function. The latter one, in turn is defined as a limit of an integral of a simple ... 9 Saying f is blah almost everywhere means, by definition, that the set of points where f is not blah has measure zero. The empty set has measure zero. So if f is blah everywhere then f is blah almost everywhere. In particular, if f:X\to\Bbb R then f(x) is finite for every x, hence f is finite almost everywhere. The notion "finite almost ... 0 Not an answer, a comment too long for the comment box: Oh that's you - diidn't notice the name. Didja notice someone posted a proof for that Fourier series thing? Heh. Don't know why you ask, when I said I hadn't read your proof I didn't mean to suggest I doubted it. Didn't have an actual proof in mind, seemed obvious... Hmm. Ok, first the continuity ... 1 Suppose \mu = \delta_0. Then we have for the set A = (0,1] and B= \{0\} that \mu(A) =0 and m(B)=0 while A \cup B = [0,1] and A \cap B = \emptyset. So we have property 1. For property 2 we have \mu ( [0,t])=1 for all 0 \le t \le 1, so clearly \mu( [0,t]) is continuous in t, thus also property 2 holds. For the last property, for any f ... 0 Here's a strong counterexample: For every X \subseteq [0, 1], there is a Y \subseteq X such that \mu^{\star}(X) = \mu^{\star}(Y) = \mu^{\star}(X \setminus Y). So letting A_1 = Y, A_2 = X, T = X \setminus Y gives you a strong counterexample. For our purpose, it is enough to take A_2 = [0, 1] and A_1 to be a Bernstein subset of [0, 1] (which ... 0 Well, you have to prove that it is closed in respect to finite union, intersection and complement. The union is pretty clear (it is the definition of M_0), the complement doesn't seem to be very difficult too... As for the intersection, I think that you should first analyse an intersection of two rectangles. And then you decompose the intersection of a ... 2 It is true that$$ m(E) = 0 \quad \iff \quad \forall \epsilon > 0 \, : \, m(E) < \epsilon. $$This has nothing to do with measure theory; it's a property of real numbers that if x \ge 0 is less than any positive number, then x = 0. However, the statement of Lusin's theorem is not this. Lusin's theorem says$$ \forall \delta > 0 \; \exists F ... 2 An example of a filtered probability space is$([0,1), \mathbf{B}_{[0,1)}, \{\mathcal{F}_t\}_{t\geq 0} , \mathbb{P})$, where, for each$t\in \mathbb{R}, t\geq 0$, define $$\mathcal{F}_t = \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right]}=\textrm{ the Borel \sigma-algebra defined in } \left [0\,,\,1-\frac{1}{t+1} \right]$$ Note that, by definition of ... 0 The condition to ensure the tightness is that$\displaystyle \frac{f(x)}{x} \to \infty$as$x \to \infty$. In fact if e.g.$f(x)=|x|$then $$\sup_n \int f \, dμ_n<\infty$$ is only a necessary condition for the tightness. 1 Indeed, the composition of measurable functions remains measurable. In your case, both,$M(t)$and$x\to x²$are measurable. 0 Let$G\in \Sigma$with$\mu(G) = \infty$. Choose$X_i \in \Sigma$with$\cup_{i=1}^\infty X_i = X$and$\mu(X_i) < \infty$for all$i$by the definition of$\sigma$-finite. Without loss of generality assume the$E_i$are increasing (replace$E_i$with$\cup_{j\leq i} E_i) Then $$G = \cup_{i=1}^\infty (G \cap X_i)$$ so $$\infty = \mu(G) = \lim_{i} ... 1 Is your question just about \sigma-algebras or about filtration?? If you are just looking for an example of an uncountable increasing family of \sigma-algebras, there are many examples. One of the simplest examples is: for each t\in \mathbb{R}, t\geq 0, define \mathcal{F}_t = \mathbf{B}_{[0,t+1]}\textrm{ the Borel \sigma-algebra defined in } ... 4 Consider the sequence defined by$$f_n = \begin{cases} \chi_{[0,1]} & n = 1\\ \chi_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} & 2^k \le n < 2^{k+1} \end{cases}$$Then \|f_n\|_1 = \frac1{2^k} \to 0 but f_n \not\to 1 on any x\in[0,1]. You can, however, prove that there exists an a.e. convergent subsequence with limit 0 in this ... 2 You can look at$$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$For r\in[0,1]. Where Borel means the standard Borel sets of [0,1]\times[0,1]. Then for r\leq s we have \cal B_r\subseteq B_s, since whenever A\in\cal B_r, we ... 4 Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. \mathcal{F}_t=\sigma(W_s:s \in [0,t]). This sort of thing is usually how filtrations are created. 0 You can do the same thing as you do in the discrete-time case by just thinking of the X_n as being indexed by the rationals. Explicitly, choose a bijection f:\mathbb{N}\to\mathbb{Q}, and for each r\in\mathbb{R}, let \mathcal{F}_r be the \sigma-algebra generated by the X_n such that f(n)<r. 3 Suppose there were such continuous maps. Then \eta would be injective. Hence it's restriction to the closed unit disc is a homeomorphism. The image of that disc would be a closed bounded interval. But the disc is not homeomorphic to an interval: removing an interior point from the interval leaves a disconnected set, while removing any point from the disc ... 4 A continuous map from \Bbb R^2 to \Bbb R cannot be injective. This is trivial; proof below. It's higher-dimensional versions, like the fact that a continuous map from \Bbb R^3 to \Bbb R^2 cannot be injective, that I suspect are non-trivial. Given two points a,b\in\Bbb R^2, let [a,b] denote the line segment from a to b. Say f:\Bbb ... 0 You are misreading him. Cohn doesn't remark \mu(P)>0 and then deduce \nu_0(X)-\epsilon \mu(X)=(\nu_0-\epsilon \mu)(N) \le 0. Rather, he proves \mu(P)>0 by showing that if, to the contrary, \mu(P)=0 were true, then he could deduce \nu_0(X)-\epsilon \mu(X)=(\nu_0-\epsilon \mu)(N) \le 0, which he has already shown to be impossible. 2 By the (multi-dimensional) Leibniz formula, we have$$\begin{align*} \partial^{\alpha} \phi_k&= \sum_{\beta+\gamma = \alpha} \underbrace{c_{\beta,\gamma} (\partial^{\beta} \phi) \cdot (\partial^{\gamma} (1-\psi(k \bullet)))}_{=:S_{\beta,\gamma}} \end{align*}for some constants c_{\beta,\gamma} (which can be calculated expliticly). If \gamma=0, ... 2 Question 19 For a complex measure \gamma and a set B, notice that \gamma(A) = 0 for every measurable A \subset B iff \lvert\gamma\rvert(B) = 0. And this happens iff \lvert\gamma\rvert(A) = 0 for every measurable A \subset B. From the above, \nu \perp \mu iff \lvert\nu\rvert \perp \lvert\mu\rvert. And, \begin{align*} \lvert\nu\rvert \ll ... 1 This solution uses the hint given in Etienne's comment. Let \limsup_{n\to\infty}A_n=A,\liminf_{n\to\infty}B_n=B,\limsup_{n\to\infty}A_n\cap B_n=E. A\cap B\subseteq E\begin{align} &x\in A\cap B\implies \color{blue}{(x\in B)}\land \color{red}{(x\in A)}\\ \implies&\color{blue}{\left(x\in\bigcup_{k=1}^\infty\bigcap_{n=k}^\infty ... 0 If the interval is[a,b]$,$\mu([a,b]) = f(b) - f(a) = 2^{-n}$. 3 Let$X$be a non-empty set and$\mathscr A$an algebra on it. A premeasure on a$\mathscr A$is a function$\lambda:\mathscr A\to[0,\infty]$such that$\lambda(\varnothing)=0$; and if$A_1,A_2,\ldots$is a countable collection of disjoint sets in$\mathscr A$and if their union is contained in$\mathscr A$, then $$\lambda\left(\bigcup_{n=1}^{\infty} ... 0 Yes. It's certainly true if \int f\,d\mu=\infty; assume then that \int f\,d\mu<\infty. Just to simplify the notation we can assume that f>0 (consider X'\subset X). And since f is integrable the set where f=\infty has measure zero, so we can also assume f<\infty. For \lambda>1 and n\in\Bbb Z let ... 2 The usual definition to make in this context is to say that the "valueless" part is the largest open set which has zero value. As you mention, there is typically no largest set which has zero value. However, there is always a largest open set of zero value: take the union of all open sets of zero value. This union will still have zero value, because it is ... 0 The definition of derivatives on the boundary of a domain is not completely agreed upon. My preferred definition of differentiability of a function f\colon A \to \mathbb R with A\subset \mathbb R^n in any point x_0\in A is the following. We say that f is differentiable in x_0 if there exists a linear map L such that$$ \lim_{x\to x_0} ... 3 The weak topology has the property that the dual is the same as the dual with respect to the original topology, if$(E,\tau)$is a topological vector space, and$E^\ast$its (topological) dual, then $$(E,\sigma(E,E^\ast))^\ast = E^\ast.$$ For the topology of pointwise convergence on$C(S)$, the dual is easily described: if$\lambda \colon C(S) \to ...

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For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$\{ x: \in \mathbb R : f(x) > a \},$$ although it is not always measurable.

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$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...

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$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...

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Given a measureable $f:(X,\mathcal{M},\mu)\to(Y,\mathcal{N})$ (a measure space to a measurable space) one can define a measure $\nu=f_{*}\mu$ on $(Y,\mathcal{N})$ by $$\nu(B)=\mu(f^{-1}(B)).$$ Both $f$ and $\mu$ are necessary to define it, so $f_{*}\mu$ seems like good notation to me.

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Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation. Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on \$(M, ...

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