# Tag Info

0

The expected number of flips can be infinite. I first show how to make it larger than any finite number. Choose a large integer $n$. Divide $[0,1]$ into $2^n$ diadic intervals, which we indexed by $W_n:=\{ 0, 1 \}^n$. Choose real numbers $\alpha < \beta$. Let $X$ be the union of those dyadic intervals indexed by $w \in W_n$ for which at most $n+ \alpha ... 1 I can sharpen PhoemueX's very nice answer a little bit. In particular, yes,$\mathbb{Q}$is a counterexample. You are quite right that, if a set$B$is to be the measure-independent limit of a sequence of sets$A_n$, we must have$1_{A_n} \to 1_B$pointwise. In particular, as PhoemueX says, we must have$B = \bigcup_{k \ge 1} \bigcap_{n \ge k} A_n$. So ... 0 You're almost done. Note that$g'(x)$dominates$\sqrt{x} g'(x)$on this interval, and the integral of$g'(x)$is less than or equal to$g(1)-g(0)$. This doesn't require absolute continuity for$g$. See Proposition 22 here. 2 I think the terminology comes from the Lebesgue case. Here when a measure is absolutely continuous with respect to Lebesgue measure, it has a density with respect to Lebesgue measure (its Radon-Nikodym derivative). This density is called an absolutely continuous function. To my understanding the notion of absolutely continuous functions is the older notion. ... 1 This is not a complete answer, but maybe helpful nevertheless: As you noted, it is a necessary condition that$\chi_{U_n}$(the characteristic function/indicator function of$U_n$) fulfils$\chi_{U_n} \to \chi_B$pointwise. But conversely, if this is the case, then $$\mu(B \Delta U_n) = \int |\chi_B - \chi_{U_n} | \, d\mu \to 0$$ by dominated ... 0 I will give a short answer for general filtrations. ad 1) Adaptedness is not enough. Let$f: [0,\infty) \to \mathbb{R}$be a non-measurable function. Set$Y = f$for all$\omega\in\Omega$. For all$t\in[0,\infty)$we have that$Y_t$is a deterministic value, hence a random variable and$\mathcal{F}_t$-measurable. So$(Y_t)$is an adapted process. ... 2 For every measurable function$u$,$Y=u(X_1,\ldots,X_n)$is$\sigma(X_1,\ldots,X_n)$-measurable since, for every Borel subset$B$,$[Y\in B]=[(X_1,\ldots,X_n)\in u^{-1}(B)]$and$u^{-1}(B)$is a Borel subset of$\mathbb R^n$. 3 Each$A_n$is a subset of$\Omega$. In the previous axiom we only needed to refer to one arbitrary set, so we just called it$A$. Now we want a collection of countably many arbitrary subsets$A_1,A_2, A_3,A_4,A_5,...$so we need to label them somehow, and we do that via subscripts running over the natural numbers. Now remember, a$\sigma$-algebra is a ... 0 Suppose${\cal S}$is a collection of sets. We can form a$\sigma$-algebra by taking$X = \cup_{A \in {\cal S}} A$, and letting${\cal F} = \{ A | A \subset X \}$which is trivially a$\sigma$-algebra. Hence there exists a$\sigma$-algebra containing the collection${\cal S}$. Now let${\cal C}_{\cal S} = \{ {\cal F} | {\cal F} \text{ is a } \sigma ...

1

Let $F_1(n,x):=\frac{(2n+1)^2}{(n(n+1))^2}$ and $F_2(n,x):=sin((2n+1)x)$. So it is enough to show that $F_1$ and $F_2$ are $\mu\times\mathcal{L}$-measurable. It is clear that $F_1$ is $\mu\times\mathcal{L}$-measurable. For the measurability of $F_2$, we consider the maps $f: [0,\pi]\to \mathbb{R}$ by $f(x)=x$ and $g:\mathbb{N}\to\mathbb{R}$ by $g(n)=2n+1$. ...

2

Hint: $X_1 + \ldots + X_n < t$ iff there exist rational numbers $r_1, \ldots, r_n$ such that $r_1 + \ldots+ r_n < t$ and all $X_i \le r_i$.

1

You can use a direct proof, rather than contradiction. Choose $\epsilon>0$ and let $A_{n,\epsilon} = \{ x | |f_n(x) -f(x)| > \epsilon \}$. Then $\int |f-f_n|^p \ge \int_{A_{n,\epsilon}} |f-f_n|^p \ge \epsilon^p \mu(A_{n,\epsilon})$, and so ${1 \over \epsilon^p}\|f-f_n\|^p \ge \mu(A_{n,\epsilon})$. It follows that $\lim_n \mu(A_{n,\epsilon}) = 0$ and ...

2

Let $E_{n,\varepsilon} = \{x : |f_n(x)-f(x)| > \varepsilon$. If $\mu(E_{n,\varepsilon}) > \delta$ then $$\int|f_n-f|^p\geqslant \int_{E_{n,\varepsilon}}|f_n-f|^p > \int_{E_{n,\varepsilon}} \varepsilon^p = \mu(E_{n,\varepsilon})\varepsilon^p>\varepsilon^p\delta,$$ and the desired inequality is obtained by raising both sides to the $\frac1p$ ...

1

As suggested in the comments, the sequence of partial sums $g_k = \sum_{n=1}^k |f_{n+1} - f_n|$ is monotonically increasing and converges to $g$. Then it follows from Monotone Convergence Theorem that $\lim \int g_k = \int g$ and similarly $\lim ||g_k||_p = \int ||g||_p$. Applying triangle inequality and $||f_{n+1} - f_n||_p < 2^{-n}$, you can show each ...

1

When you are talking about sets, and you say the sequence is "increasing" or "non-decreasing", it just means that you have the containment $A_{1} \subset A_{2} \subset A_{3} \subset \dots \subset A_{n} \subset A_{n + 1} \subset \dots$ (usually, these are all proper subsets if you say "increasing" while they would not all necessarily have to be proper if you ...

3

It means that they satisfy $A_n \subset A_{n+1}$. The limit is not a limit in the $\epsilon$-$\delta$ sense. It just means $\cup_{n=1}^\infty A_n$. For example, take $A_n = [0,n]$, then $\cup_{n=1}^\infty A_n = [0, \infty)$. From a probability perspective, there is a real limit associated with these nested sets. Suppose $p$ is the probability measure. ...

1

Don't integrate over the whole line. Integrate over the interval $[-M,M]$: $$\int_{[-M,M]} \sum_{k=1}^\infty \frac{a_k}{\sqrt{|x - \alpha_k|}} \, dx = \sum_{k=1}^\infty a_k \int_{[-M,M]} \frac{1}{\sqrt{|x - \alpha_k|}}\, dx.$$ The last integral does not exceed $4 \sqrt{M}$, so that the series has finite integral on $[-M,M]$, and is thus finite almost ...

1

The presence of the $\cos$ term leads one to think that there might be some cancellation in the integral and that the triangle inequality is not the right way to start. For a prelim question that is a bit misleading, in my opinion. For any $y$ you have $$\left| \int_0^\infty \int_t^\infty f(x,y) \cos \left( \frac ty \right) \, dx dt \right| \le ... 0 \int_{0}^{\infty}\int_{t}^{\infty}... dx dt=\int_{0}^{\infty}\int_{0}^{x}... dt dx. Thus \int_{0}^{\infty}\int_{0}^{x} f(x,y) \cos(\frac {t}{y}) dt dx=\int_{0}^{\infty} f(x,y) y\sin(\frac{x}{y})dx. Because \sin(x)\leq x for  x\geq 0 we have \int_{0}^{\infty} f(x,y) y\sin(\frac{x}{y})dx\leq \int_{0}^{\infty} xf(x,y) dx. now you can put the absolute ... 1 Use Urysohn to construct g with g \equiv 2 on U (multiply your g by 2). Then take a sequence (f_n) withf_n \to g uniformly. For n large, you will then have f_n \geq 1 on U. 0 Sam Drury's class notes for his Math 355 course at McGill contains a clever proof of this under Section 4.6, "Infinite products of probability spaces". I'm not sure I know how to give proper hints based on it, but I hope the outline below is a decent approximation. Drury first makes the following definitions for convenience: A set C \subset ... 0 No, you can't use the same a. Let \Phi_\alpha = \{ x | |f(x)| > \alpha \}, \Gamma_\beta = \{ x | |g(x)| > \beta \} Let \alpha  be such that \mu \Phi_\alpha = 0, and \beta  be such that \mu \Gamma_\beta = 0 and suppose x \in (\Phi_\alpha \cup \Gamma_\beta)^c . Then |f(x)+g(x)| \le \alpha + \beta, and since \mu \{x | |f(x)+g(x)| ... 2 We have |f(x) + g(x)| \leq |f(x)| + |g(x)| for every x, by the triangle inequality. Now |f(x)| \leq \|f\|_\infty almost everywhere, and similarly |g(x)| \leq \|g\|_\infty almost everywhere. Therefore,$$|f(x) + g(x)| \leq \|f\|_\infty + \|g\|_\infty$$almost everywhere. Since the right hand side is an almost-everywhere upper bound for |f(x) + ... 1 A set of points on the x -axis is said to have measure zero if the sum of the lengths of intervals enclosing each of the points can be made arbitrarily small. If f(x) is bounded in [a,b] , then a necessary and sufficient condition for the existence of ∫ b a f(x)dx is that the set of discontinuities have measure zero. Great clarification results by ... 1 angryavian's answer should also point you into the right direction for your second question on why no norm-minimal element exists: For any f \in C there holds$$\int_0^{1/2} f(x) \,\text{d}x \leq \frac 1 2 \| f \|_\infty \quad\text{and}\quad - \int_{1/2}^1 f(x) \,\text{d}x \leq \frac 1 2 \| f \|_\infty,$$so a norm-minimal f, i.e. \| f \|_\infty = 1/2 ... 3 In my comment above, I noted$$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt} \le \|f\|_\infty.$$I have a hunch that the definition of M should be$$\int_0^{1/2} f(t) \mathop{dt} - \int_{1/2}^1 f(t) \mathop{dt}=1/2$$in order for part ii) to work. [Or more generally, the number in the definition of M and the number in part ii) should be ... 3 Here is an important thing to keep in mind: If you want to show that the function$$ \Bbb{R} \to \Bbb{R}, x \mapsto e^{x^2 \cdot \sin(x)} - \sum_{n=0}^\infty \frac{x^n}{(n!)^2} $$is continuous, you will (hopefully) not try to find for every \varepsilon > 0 some \delta > 0 such that ... Instead, you will rely on certain closure properties of ... 0 As noted by copper.hat's answer, the result is theorem 1.40 on Rudin's book. For anyone who should stumble upon this question, I place here a screenshot of the proof. I think this does not violate any kind of copyright, since the book can anyway be fully downloaded from the above provided link. 1 Suppose that s is a step function. Then for any a \in \mathbb R the set \{s > a\} = \{x \in \mathbb R : s(x) > a\} is either empty or a finite union of intervals. Since all intervals are F_\sigma sets, so is \{s > a\}. Now suppose that s_n \nearrow f. Then for any a \in \mathbb R you have \{f > a\} = \bigcup_n \{s_n > a\} so ... 3 Let f: [0,1] \to [0, \infty) be the characteristic function of the set of irrationals in [0, 1]. Assume there exists a nondecreasing sequence of step functions \{s_n\} converging pointwise to f. Without loss of generality we can assume that n \ge 0. Since rationals are dense and \{s_n\} is a nondecreasing sequence of step functions, s_n(x) ... 2 Let f:[0,1]\to[0,\infty) be given by$$f(x)=\begin{cases} 0 &; x\in\mathbb{Q} \\ 1 &; x\notin\mathbb{Q}\end{cases}$$If 0\leq s\leq f is a step function, then s is non-zero for at most finitely many points. Hence, any well-defined limit \lim_{n\to\infty}s_n of step functions 0\leq s_n\leq f is non-zero for at most countably many points and ... 1 This is exactly Theorem 1.40 in Rudin's "Real & complex analysis" (3rd ed.) 2 For the first part, since f is bounded:$$ |F(x)| = \left|\int_{\mathbb R} K(x-s)f(s)\,ds\right| = \left| \int_{\mathbb R} f(x-s)K(s)\,ds \right| \le M \int_\mathbb R |K(s)|\,ds where M = \sup_\mathbb{R} |f|. (The absolute values are missing in @Ant's solution.) For the second part, \begin{align} \lim_{x \to \pm\infty} F(x) &= \lim_{x \to ... 1 Note that f is bounded on \mathbb R, so set M = \sup |f(x)|, thenF(x) = \int_\mathbb R K(x-s)f(s)ds = \int_\mathbb Rf(x-s)K(s)ds\le M \int_\mathbb R K(s)ds$$and since K \in \mathcal L^1(\mathbb R) you are done. To show that \lim_{x \to \infty} F(x) = 0, note that for every \varepsilon > 0, exists N > 0 such that for every x > ... 1 First of all, your second and third paragraphs. Recall that if \xi is F-measurable and F\subseteq G then \xi is G-measurable: that follows from the definition of measurability. Now, if \Bbb F = (F_n)_{n\geq 0} is a filtration and \xi = (\xi_n)_{n\geq 0} is \Bbb F-adapted, then for each n we have \xi_n is F_n-measurable. Since ... 1 It just as simple as this: Of course \mathcal{H}^1(\Gamma\cap A) \leq \mathcal{H}^1(A) and since \dim_H A <1 you have \mathcal{H}^1(A) = 0. 0 I do not know if following solution is correct or not. However I am also attempting to understand this problem and did not want to replicate this question. Please guide me towards correct solution. We know that the set of integers has same cardinality as set of rational numbers. Specifically according to Cantor's scheme: We list all rational numbers for ... 0 Definition(Measurable function): Let f be a function from a measurable space (\Omega , \mathcal F) into the real numbers. We say that the function f is measurable if for each Borel set B \in \mathcal B , the set \{\omega; f(\omega) \in B\} ∈ \mathcal F. Definition( random variable): A random variable X is a measurable function from a probability ... 0 (2) \implies (3): Let \varepsilon>0 be given, and let A\subset E\subset B so that A and B are elementary and m(B-A)<\varepsilon. Since A\subset E\subset B and by monotonicity, we have:$$m^*(E\triangle B)=m^*(B-E)\le m^*(B-A)<\varepsilon.$$For the implication (3)\implies (1) we need the following lemma: If B is Jordan ... 1 Due to the completely different nature of the Riemann and Lebesgue approaches, there is no direct proof of the equivalence you mention as far as I know. To show that the class of absolutely HK-integrable functions is the class of Lebesgue integrable functions, you use the following characterisation: If V is a vector space of Lebesgue measurable ... 3 Use the Dominated convergence theorem.  f_ n converges pointwise to 0 and is bounded by 1/\sqrt x which is integrable. 1 It seems the following. :-) Suppose that B is not Jordan measurable and A=B. Then$$m^*(B-A)=m^*(B-B)=m^*(\varnothing)=0<m^*(B)-m_*(B)=m^*(B)-m_*(A).$$1 Haven't received any answers, so here's my take after some additional reflection: the answer is yes - partly. Some aspects are generalized, mostly those dealing with integration over spaces, as tangentially/indirectly discussed in the preface of Real Analysis: Modern Techniques and Their Applications by Folland. 0 @WillJagy pointed me to Geometric Measure Theory, which requires measures to understand areas/lengths of things like fractals. 1 The expectation \mathbb{E}[Y_n] is equal to 1 for all n; the probability \mathbb{P}\{Y_n > 0\} = \frac{1}{2^n} goes to 0. However, note that with probability 1/2^n (i.e., when Y_n is not 0), Y_n = 2^n, which itself goes to infinity. The expectation is fixed and equal to 1, the probability p_n to be non-zero becomes smaller and ... 3 A function is absolutely continuous if and only if it is an antiderivative of some integrable function (this is the approach via measure theory, which I assume is what you want: it connects to the property \ m(A) < \delta \Rightarrow \int_{A} f < \epsilon that you mentioned). Given g(x)=x|\sin(x^{-1/2})|, you can find it's differentiable ... 3 Let A = A_0 \cup A_1, where A_0 = \{x\in X : \lim f_n(x) < \infty\} and A_1 = \{x\in X : \lim f_n(x) = \infty\}. Then x \in A_0 if and only if for every j\in \Bbb N, there exists an N \in \Bbb N such that |f_m(x) - f_n(x)| < \frac{1}{j} for all m, n \ge N. Thus$$A_0 = \bigcap_{j = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{m = N}^\infty ...

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Have $\mu$ be a $\sigma$-finite measure on the $\sigma$-algebra $\mathscr{M}$. Have $\mu^*$ be the outer measure induced by $\mu$, $\mathscr{M}^*$ the $\mu^*$-measurable sets and $\overline{\mu}$ the measure $\mu^*$ restricted to $\mathscr{M}^*$. We want to prove $\overline{\mu}$ on $\mathscr{M}^*$ is the completion of the measure $\mu$ defined on ...

6

I will show that $\mathscr M^*=\overline{\mathscr M}$. From Theorem 1.9 and Carathéodory's theorem it follows, indeed, that $\mu^*|\mathscr M^*$ is the completion of $\mu$, since $\mu^*|\mathscr M^*$ is a complete measure on the $\sigma$-algebra $\mathscr M^*=\overline{\mathscr M}$ and it extends $\mu$ by Proposition 1.13. Consider a measure space ...

1

In all fairness, the formula $\displaystyle\int x^{-1}dx=\log |x|+c$ is incorrect, or at least is unexhaustive. The correct formula is $$=\begin{cases}\log x+c_1;x>0\\\log(-x)+c_2;x<0\end{cases}$$ where $c_1,c_2$ are arbitrary constants. The point is that the singular point at $x=0$ makes the solution space of $xy'=1$ two dimensional instead of $1$ ...

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