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0

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 ...


1

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 ...


1

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} ...


1

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 ...


1

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 $f_{n_{k_h}}$ ...


2

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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