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First, a few words about inverse images. If $X$ is a set, then the powerset $\mathcal{P} X$ is a poset. If $f: X \to Y$ is a function, then the inverse image function $\mathcal{P}Y \to \mathcal{P}X$ is order-preserving, and it has both left and right adjoints -- which in particular means it preserves both meets and joins. The left adjoint is the ...

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The coin tosses can be modelled as a sequence of independent $\text{Ber}\left(\frac{1}{2}\right)$ random variables $(X_n)_{n\in\mathbb{N}}$ Letting $A_n=\{X_n=1\wedge X_{n+1}=1\}$, you want to know $P(\lim\sup_n A_n)$ Consider $\lim\sup_nA_{2n}\subseteq\lim\sup_nA_n$. Notice that $(A_{2n})_{n\in\mathbb{N}}$ are independent. Since ...

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Firstly, just to clarify, by $\mu\leq \nu$ I expect you mean $\mu(A)\leq \nu(A)$ for $A$. We therefore have that $\mu$ is absolutely continuous wrt $\nu$ (written $\mu\ll\nu$, and why I'm asking about the meaning of your inequality) and so there is a function in $L^1(\nu)$, $\frac{\mathrm d\mu}{\mathrm d\nu}$ such that $$\int_E \frac{\mathrm d\mu}{\mathrm ... 4 Consider using Holder's inequality with conjugate exponents p and p/(p-1):$$\left|\int_{A_n} f\, d\mu\right| \le \int_X \chi_{A_n}|f|\, d\mu \le \|\chi_{A_n}\|_{p/(p-1)} \|f\|_p = \mu(A_n)^{(p-1)/p} \|f\|_p.$$Since \mu(A_n) \to 0, the result follows. 0 We have to expect that the anwser will depend on a. You have a little problem with your answer when a=1. Treat that case separately (it converges). For a>0,a\ne 1, what you did is fine. You arrive at a one-variable integral where the integrand appears to have two singularities, at 0 and 1. But the singularity at 0 is removable. At 1, the ... 2 An example. F = [0,7] with its usual metric. Then: \mathcal H^s(F) = 0  for s>1, \mathcal H^1(F) = 7, \mathcal H^s(F) = \infty for 0<s<1. Thus, according to your definition, \dim F = \inf\;(1,\infty) = 1. 2 \dim F is not defined as a value for which the Hausdorff dimension equals 0. It's defined as the infimum of a set of such values, which does not mean it has to be a member of the set itself. 1 In general, we can hardly expect to find F,G such that F \subseteq A \subseteq E \subseteq G and |G \backslash F| < \epsilon (as |E \backslash A|>\epsilon for \epsilon sufficiently small). Hints: Using Urysohn's lemma, prove that the claim holds true for E = \mathbb{R}^d and A open. (That's basically what you already did). Conclude ... 0 If I'm interpreting this correctly, what they mean is this: Suppose f has a discontinuity at some y\in[a,b]. Since f is right-continuous, f_y:=\lim_{x\rightarrow y^+}f(x) exists. Since f has at most countably many discontinuities, we may assume w.l.o.g. that f(y)=f_y. Here, the "upper value" is f_y, which, as f is increasing, satisfies ... 1 For the first part: Define f_n = 1_{[0,1]} for all n (this is a constant sequence), and let f=1_{[1,2]}. Then \int f_n =1 = \int f for all n, but \int |f_n - f| =2 for all n. For the second part: Take some enumeration \{I_n\} of all subintervals of [0,1] which have the form [\frac{k}{m},\frac{k+1}{m}] for some integers m,k. Let ... 0 Some examples on [0,1]: For the first one, let f_n(x)=\cos nx, f(x) = 0. For the second one, consider the characteristic functions, in order, of [0,1],[0,1/2],[1/2,1],[0,1/3],[1/3,2/3],[2/3,1], \dots. 1 The following two counterexamples take place in  [0,1]  with its standard  \sigma -algebra and Lebesgue measure. Counterexample for the first: Let  (f_{n})_{n \in \Bbb{N}} \stackrel{\text{df}}{=} \left( \chi_{[n,n + 1]} \right)_{n \in \Bbb{N}}  and  f \stackrel{\text{df}}{=} \chi_{[0,1]} . Counterexample for the second: Let  H_{n}  denote the ... 0 It looks like on the second slide of page 2 on this PDF file, "All σ-algebras are algebras, and all algebras are semi-rings." 0 I will assume you're working in \mathbb{R}. The definition of outer-measure is here:$$m^{*}(E)=\inf \{\sum_{n=1}^{\infty}l(I_n)| E \subset \bigcup_{n=1}^{\infty}I_n, I_n \text{being disjoint sequence of open intervals}\}$$If E has outer measure zero, that means that for every positive \epsilon, there exists a sequence of disjoint open intervals ... 0 I worked on this for a couple hours and think I have come up with a solution. I found it is easier to prove that B^c is closed: Suppose \sup_{C_r \in \mathcal{C}_r(x_n)} \frac{\mu(C_r)}{m(C_r)} \leq a for x_n \to x. Let C_{k} \in \mathcal{C}_r(x) be a sequence of sets such that \frac{\mu(C_{k})}{m(C_{k})} \to \sup_{C_r \in \mathcal{C}_r(x)} ... 3 Lets consider \Omega=\{1,2,3,4\} \sigma(\{1\}) is the smallest \sigma-algebra which contains 1. So we must take any other elements of P(\Omega) such that the conditions for being a \sigma-algebra are fulfilled. It does clearly contains 1. Also 1^C=\{2,3,4\}. And \Omega,\emptyset. So we have ... 2 Say we are doing this on the real line \mathbb R. Let \mathcal G be the collection of all open sets. We are interested in \sigma-algebras \mathcal F such that \mathcal F \supseteq \mathcal G. There may be many such \sigma-algebras. For example, the power set \mathcal P, consisting of all subsets of \mathbb R is one. But that is the ... 1 if you agree that the borel \sigma-algebra isn't neccessarily all the subsets of the topological space, then you might also agree that P(X) (or 2^X in a different notation, the power set) is a larger \sigma-algebra. Larger means more sets in the \sigma-algebra. Also note that the generated \sigma-algebra by the open sets, which is the borel ... 1 By translating we may assume x=0 for convenience, and then by rescaling we have$$\frac{1}{c_nr^n}\lambda_n(B_r(0)\setminus B_r(y)) = \frac{1}{c_n} \lambda_n(B_1(0)\setminus B_1(y/r)).$$Now observe y/r\to 0 as r\to\infty and say "dominated convergence theorem". The dominated convergence theorem is overkill. To be more elementary, B_1(0)\cap ... 4 No, this isn't true. As an example:$$f(x)=\begin{cases}1 &\text{if } x\in (0,1/2] \\ 0&\text{else}\end{cases}$$and$$g(x)=\begin{cases}1 &\text{if } x\in (1/2,1) \\ 0&\text{else}\end{cases}$$3 Counterexample: Consider (0,1) with Lebesgue measure and the characteristic functions \chi_{(0,\frac{1}{2})} and \chi_{(\frac{1}{2},1)}. Then \chi_{(0,\frac{1}{2})}\chi_{(\frac{1}{2},1)} = 0 but the product of their integrals is \frac{1}{4}. 1 In addition to the ones mentioned above, there are also (in no particular order): R. Bartle, Introduction to Measure theory - it has a particularly nice section on integrals of the form$$ h(x) = \int f(x,t)dt $$G. De Barra, Measure Theory and Integration"- does a nice job of differentiation. Halmos, Measure Theory - a little dated now, but I have looked ... 1 The 3 big options are: Folland, Real Analysis and its Applications 2e Royden & Fitzpatrick, Real Analysis 4e Rudin, Real & Complex Analysis 3e I liked each of them for different reasons, but I found Folland + Royden & Fitzpatrick (To supplement chapter 3 of Folland) was a good option. Lots of solved problems is generally an undergraduate ... 0 I am keen on the Schaum's outline series. Try this: "Schaum's Outline of Theory and Problems of real Variables: Lebesgue Measure and Integration with Applications to Fourier Series" by Murray Spiegel This series of books have hundreds of solved problems on measure theory (& other topics like topology, complex analysis & differential geometry). I ... 1 If f_n\in L^1, then necessarily there exists C_n>0 such that \|f_n\|_1\leqslant C_n (since by definition, each f_n has a finite L^1-norm). If there exists C>0 such that \|f_n\|_1\leqslant C for all n, then we say that f_n is uniformly bounded. 3 I'm not sure you'll be able to find a text with solved exercises. My personal favorite is Folland's Real Analysis. 1 No. Let f_n = n^3 \chi_{(0,1/n)}(x). Then f_n(x) \to 0 for every x \in (0,1) but \|f_n\|_1 = n^2. 1 It does not. For instance, f_n=n^21_{(0,\frac{1}{n}]} is a counterexample. 2 Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let \mu be a measure. It is clear that if \mu is countably additive then it is finitely additive and countably sub-additive. Assume that \mu is countably sub-additive and finitely additive. Consider a collection \{A_n \}_{n=1}^\infty of ... 0 Here is a brute force approach with nothing elegant: Let E = \{x \in X : \lim f_n(x)\; exists and is finite\}. Let f=\limsup f_n. Notice that x \in E if and only if f_n(x) \to f(x). Letting h_n := f-f_n, we see that h_n is measurable and that E = \{x \in X: h_n(x) \to 0 \}. By the definition of a convergent sequence, we know that h_n(x) ... 0 I'm sure that any subset of \mathbb R does exist its Vitali Cover(Each element of Vitali Cover, the interval, is closed). For each x ∈ X ⊂ \mathbb R, let I^{x}_{m} = [x, x+1/m] for ∀ positive integer m and let C = {I^{x}_{m}: x∈X and m=1,2,3,...}. Then C forms a Vitali cover for X. 0 the prove is clear with definition of algebra and topology . for example consider X=\Bbb{N} and$$S=\{A\subset \Bbb{N} |A\, or A^c is finite\}$$it is clear that S is algebra ,and S is not topology,because let A_{n} = \{2n\} and (\bigcup_{n=1}^{\infty}A_{n}=even numbers) is not belong to S because itself and complement of it is not finite. 3 The function f is measurable if and only if there exists a sequence of step functions that converge to f almost everywhere. Your u is integrable, hence measurable. Therefore you have that sequence of step functions with support in (0,1) which converges to u a.e. Note that the same sequence converges a.e. to \tilde u on (0,\infty), hence ... 2 Look at the inverse image of a generating measurable set.$$\bar{u}^{-1}(a,b)$$If a<0<b then because inverse images and unions commute, i.e. f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B) we get:$$\bar{u}^{-1}(a,b)=\bar{u}^{-1}(a,0)\cup \bar{u}^{-1}(\{0\})\cup\bar{u}^{-1}(0,b)=u^{-1}(a,0)\cup (1,\infty)\cup u^{-1}(\{0\})\cup u^{-1}(0,b)$$... 0 Let S denote the class of sets under consideration. All you need to show is that if A_n \in S for all n \in \mathbb{N} then A:= \bigcup A_n is in S too. So define$$B_n:= \bigcup_{k \leq n} A_k. $$The B_k form an increasing sequence, and they increase to A which is thus in S (by the monotone class property). 1 Hints: Let \arg\min_{{b_0,b_{-0}}}E\left[(X_{n+1}-b_0-b_{-0}'X)^2\right]=(\beta_0,\beta_{-0}')'=\beta\in\mathbb{R}^{n+1}, \quad X=(X_1,\dots,X_n)'. \beta_0=E[X_{n+1}]-\beta_{-0}'E[X], \quad \beta_{-0}=Var(X)^{-1}Cov(X,X_{n+1}). Cov(X, X_{n+1}-\beta_0-\beta_{-0}'X)=0. Use normality of (X_1,\dots,X_{n+1})'. Show that ... 1 HINT: Use Fubini's theorem and the fact that the double integral is zero for all a and b to show that the integral of the function is 0 on every rectangle in [0,1]\times [0,1]. Then prove that any function that is not almost everywhere 0 must have nonzero integral on some rectangle. 1 Let F(x)=\int_0^x f. Then F\equiv 0 from the given hypothesis. Therefore F'\equiv 0. But F'(x) = f(x) for a.e. x by the Legesgue differentiation theorem. Thus f=0 a.e., hence \int_E f = 0 for any measurable set E. (Using a big gun there, but thought I'd toss this in.) 1 Hint: We have,$$f = f^+-f^-$$where f^+ and f^- denote the positive and negative part of f, respectively. By assumption, the (\sigma-finite) measures$$\nu(dx) := f^+(x) \, dx \qquad \mu(dx) := f^-(x) \, dx$$satisfy$$\mu((a,b)) = \nu((a,b)).$$Conclude from the uniqueness of measure theorem that \mu = \nu on \mathcal{B}(\mathbb{R}). 0 Hint: We can find a countable collection of open intervals I_k such that$$ E \subset U = \bigcup_{k \in \Bbb N} I_k, \qquad \left(\sum_{k=1}^\infty m(I_k) \right) - m(E) < \epsilon $$Now, note that$$ \left| \int_U f\,dx - \int_E f\,dx \right| = \left| \int_{U \setminus E} f\,dx \right| $$Alternative: Show that \int_U f\,dx = 0 whenever U ... 5 Hint:$$ \max(a,b)=\frac{a+b+|a-b|}{2},\quad a,b\in\mathbb{R}. 1 Lebesgue integral is first defined for non-negative functions. Then the definition is extended to general functions without sign constraint. In particular, writing the function as f=f^+-f^-, one defines \int f=\int f^+-\int f^-, where f^+ and f^- are positive and negative parts which are non-negative functions. So for Lebesgue integral to make sense, ... 0 By using the integrals \begin{align} \int \frac{x^2}{(x^2 + y^2)^2} \, dx &= \frac{1}{2y} \, \tan^{-1}\left(\frac{x}{y}\right) - \frac{x}{x^2 + y^2} \\ \int \frac{1}{(x^2 + y^2)^2} \, dx &= \frac{1}{2 y^3} \left( \frac{xy}{x^2 + y^2} + \tan^{-1}\left( \frac{x}{y} \right) \right) \\ \int \frac{y^2}{(x^2 + y^2)^2} \, dy &= \frac{1}{2x} \, ... 5 Just compute the integrals :). What is meant by this: For fixed y, we have \begin{align*} \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx &= \left[ -\frac{x}{x^2 + y^2}\right]_{x=0}^1\\ &= -\frac{1}{1+y^2} \end{align*} Hence \begin{align*} \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2}\, dx\, dy &= -\int_0^1 \frac{1}{1+y^2}\, dy\\ &= ... 1 Hints: Let x=r\cos \theta and y=r\sin \theta. 1 Yes. By definition,\sigma(\mathscr A)=\bigcap\{\mathscr B\,|\,\mathscr B\text{ is a $\sigma$-algebra and }\mathscr A\subseteq\mathscr B\},$$which can be shown to be the smallest (in the sense of set inclusion) \sigma-algebra containing \mathscr A, so that \mathscr A\subseteq\mathscr\sigma(\mathscr A). But \mathscr A is already a \sigma-algebra ... 2 By way of contradiction, assume that an integrable function  f: \mathbb{R} \to [0,\infty)  exists such that$$ \mu(E) = \mu(f^{\leftarrow}[E]) $$for any Lebesgue-measurable subset  E  of  [0,\infty) . Then$$ \forall n \in \mathbb{N}: \quad \mu({f^{\leftarrow}}[n,n + 1)) = \mu([n,n + 1)) = 1.  Hence, \begin{align} ...

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If $\mu^*$ is a metric outer measure, then all Borel Sets will be $\mu^*$-measurable, but if we have no restriction to $\mu^*$, then we can say nothing about Borel Sets. For instance, take $\mu^*$ the outer measure giving rise to the Lebesgue measure in $\mathbb{R}$, but take in $\mathbb{R}$ the discrete topology (which is a metric topology with distances 0 ...

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As of the first question, this characterisation Let $\mu$ be a finite measure. A sequence $f_n$ converges to $f$ in measure with respect to $\mu$ if and only if any subsequence $f_{n_k}$ admits a sub-subsequence $f_{n_{k_h}}$ that converges to $f$ almost everywhere. provides far more than a solid path. It's almost a proof, because the $f_{n_{k_h}}$ ...

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I believe that $\{\tau\leq 1\}$ is a shorthand for \begin{align*} \{\omega\in\Omega\,|\,\tau(\omega)\leq 1\}. \end{align*} To see that this is consistent with the statement that $\{\tau\leq 1\}=\{1\}$, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq ...

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