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10

Let $\mu$ be counting measure on $\mathbb{Z}$; that is, $$\mu(E) = |E \cap \mathbb{Z}|$$ It's easy to verify this is a measure which is not absolutely continuous with respect to Lebesgue measure; furthermore, $$\mu([-a,a]) = 1 + 2 \lfloor a \rfloor$$ (by symmetry, together with $0$ in the middle), so that $$\limsup_{a \to \infty} \frac{a}{\mu([-a,a])} = ...


7

Any countable set is a null set, there are many nonclosed countable sets. eg $\mathbb{Q}^n$


5

The set $\{\frac1n\}_{n\geq 1}$ is a nice countable (and therefore zero-measure) set in $\mathbb{R}$ that isn't closed, because it doesn't contain its limit point $0$.


5

Let $f(x) = \int_{(-\infty, x]} 1_E$. The function $f$ is continuous, $\lim_{x \to -\infty} = 0$ and $\lim_{x \to +\infty} = 1$. The intermediate value theorem does the rest.


4

$$ \left\{ \frac 1 n : n=1,2,3,\ldots \right\} $$


4

Rewrite the integral as follows $$\int_0^1 |f(x)|^p|g(x)|dx = \int_0^1 \int_0^{|f(x)|} pt^{p-1}|g(x)| dt dx$$ Switching the order of integration we obtain $$\int_0^\infty \int_{|f(x)|>t} pt^{p-1}|g(x)| dx dt = \int_0^\infty pt^{p-1}\int_{|f(x)|>t} |g(x)| dx dt$$ Now note that $\int_{|f(x)|>t} |g(x)| dx \leq \min(\frac{3}{t^2},|g|_{L^1})$ Hence the ...


4

Let $$f(x) = \sum_{n=1}^\infty (1-4^n|x-2^{-n}|)^+$$ where $a^+=\max(a,0)$. This function consists of triangular spikes of height $1$ at each $x=2^{-n}$, and the width of the spike is $4^{-n}$. It is continuous on $\mathbb R\setminus \{0\}$, but not at $0$. However, $0$ is a Lebesgue point because $$ \int_{-r}^r f(x)\,dx \le \sum_{2^{-n} \le 2r } 4^{-n} ...


4

Since $$\sin\left(\frac x n\right)\sim_\infty \frac xn$$ then we have $$ \forall x\in[0,\pi],\quad n \sin\left(\frac x n\right)f(x)\xrightarrow{n\to\infty} xf(x)$$ moreover using that $|\sin x|\le |x|,\;\forall x$ we have $$\left|n\sin\left(\frac x n\right)f(x)\right|\le |xf(x)|\le \pi|f(x)|\in L^1[0,\pi]$$ so by the dominated convergence theorem we have ...


3

As you write correctly, $n\sin(x/n) \to x$, pointwise. Now noting that $\sin x \le x$ on $[0,\pi]$, we have $n\sin(x/n) \le n \cdot x/n = x$. As $\pi|f|$ is integrable and majorizes $f(x)\sin(x/n)n$ on $[0,\pi]$, we have $$ \int_0^\pi n\sin(n^{-1}x)f(x) \,dx \to \int_0^\pi xf(x)\, dx $$ by dominated convergence.


3

A good anotated list of textbooks on geometric measure theory can be found in this blog post. Besides comments on Federer and Mattila it has several more examples. As my personal favorite I found, while lecturing geometric measure theory, "Measure Theory and Fine Properties of Functions" by Evans and Gariepy. It is short and crisp (often you have to build ...


3

I remember being a bit surprised in Rudin's Real and Complex Analysis that complex measures are by definition finite, unlike the positive measures he begins by studying. The Comment that struck you is surely meant to highlight this distinction. I too wondered "why" this distinction is necessary. My view is that it comes with having signed measures, since ...


3

Why not? Let $X=[0,1]$ and define $$\mu(E)=\begin{cases}|E|&\text{if $|E|$ is finite}\\\infty&\text{otherwise}\end{cases}$$


2

As $\nu$ is finite and Radon, there is for each $n$ a compact subset $K_n$ such that $\nu(X\setminus K_n) <1/n$. Note that for $M := \bigcup K_n$, we have $\nu(X\setminus M) =0$ and that $M$ is $\sigma$ finite for $\mu$, as it is $\sigma$ compact. Apply the ordinary Radon Nikodym theorem on $M$ and extend the Radon Nikodym derivative by zero on ...


2

Since $x^{1/n}\geq x^{1/2}$ for $n\geq 2$ and $0<x\leq 1$ and (trivially) $(1+x/n)^n \geq 1$ for $n\geq 1$, so $g(x)=x^{-1/2}$ will work fine as a dominating function. By Dominated convergence theorem, the limit thus equals $\int_0^1 e^{-x}\,dx=1-e^{-1}$.


2

Let consider a mapping $T:C(R)\to R^{Q}$ defined by $T(f)=(f(q))_{q \in Q}$, where $Q$ denotes a set of all rational numbers of $R$. It is obvious that $T$ is injective which implies that $card(C(R))\le card(R^Q)=(2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\times \aleph_0}=2^{\aleph_0}$, where $\aleph_0$ denotes the cardinality of all natural numbers. If we ...


2

Here's a direct argument following tomasz's suggestion. Let's take as given your characterization of the measurable sets $\mathcal{B}(\mathbb{R}^\mathbb{R})$. Let $A \in \mathcal{B}(\mathbb{R}^\mathbb{R})$ be arbitrary. We will show $A \ne C(\mathbb{R})$. If $A = \emptyset$ we are done. Otherwise, suppose $f \in A$. If $f$ is not continuous, we are ...


2

I'll use notation $\Delta u$ instead of $-\mu$, because it fits better in the formulas. Take a smooth domain $G$ such that $\operatorname{supp} \Delta u\subset G \Subset \Omega$. Using a smooth cutoff function, write $\psi$ as $\psi_1+\psi_2$ where $\psi_1$ has compact support in $\Omega$ and $\psi_2\equiv 0$ on $\overline{G}$. Apply Green's identity to ...


2

The condition means that, for every $n$, there exists some measurable set $A_n$ such that $$\tau\leqslant n\iff(X_1,\ldots,X_n)\in A_n.$$ Thus, every positive stopping time $\tau$ can be rewritten as $$\tau=\inf\{n\geqslant1\mid(X_1,\ldots,X_n)\in A_n\},$$ for some well-chosen sequence $(A_n)$.


2

It is correct and you use the the disjointness when you say $$ \bigcup_{k \ge n+1} A_k = A\setminus (A_1 \cup\cdots \cup A_n) $$ If the $ A_i $s are not disjoint, we only have $\supseteq $.


2

If you mean the pointwise limit inferior, then in this case it is even simpler than that: $\lim_{n\rightarrow\infty} X_n = 0$ pointwise, which one proves as you did: let $\omega$ be given and pick $N > \omega + 2$. Then, for all $n \geq N$, $\omega < \lfloor N \rfloor$ and so $X_n = 1_{[n,n+1]}(\omega) = 0$. Since this holds for all sufficiently large ...


2

The set of members of the Cantor set that are not endpoints of the deleted intervals is an uncountable set of measure $0$ that is not closed.


2

Total variation norm is a norm on measures whereas all the other types of convergence/norms/topologies that you mentioned are on spaces of functions. Nonetheless functions can be measures too. If you have a sequence of measures $\mu_n$ and say all of them are absolutely continuous with respect to a third measure $\lambda$, then the measures converge in TV ...


2

A sigma-algebra is trivial when it contains only sets of measure $0$ or $1$. There is no "0 option" here, that "we should take".


1

Fix $\beta>0$ and $c \geq 1$. For any measurable function $f(x)$ that satisfies $1 \leq f(x) \leq c$ for all $x \in \mathbb{R}$, the measure $\mu(A) = \int_{x\in A} \beta f(x)dx$ satisfies H1. Below it is shown that this example structure captures every measure $\mu(A)$ that satisfies H1 together with the properties $\mu([0,1])<\infty$ and ...


1

a.) take $A_{i}=[\frac{-1}{i},\frac{1}{i}]$. We see that $$\bigcup_{i=1}^{\infty}A_{i}=[-1,1]=A$$ From this we have that $$\bigcup_{k\geq n+1}^{\infty}A_{k}=[\frac{-1}{n+1},\frac{1}{n+1}]$$ but [ $$A\backslash(A_{1}\cup...\cup A_{n})=A\backslash A=\emptyset$$ which is do to fact $A_{1}=[-1,1]$


1

For $a)$ you want to show that the whole expression is less than than $n c^n$, where $c=\sin(a)$. In $b)$, take $u=\sin(x)$ to turn the problem into $$\lim_{n \to \infty} \int_0^1 n u^n g \left ( \arcsin(u) \right ) du$$ Now the idea is to split the region of integration into $[0,1-\delta]$ and $[1-\delta,1]$ for an appropriate $\delta$, use uniform ...


1

To expand on Did's comment: By the central limit theorem, $\frac{S_n}{\sqrt{n}} \to N(0,1)$ in distribution. Now $\frac{1}{\sqrt{\log\log n}} \to 0$ (either as real numbers, or as constant random variables converging in probability), so by Slutsky's theorem $\frac{S_n}{\sqrt{n \log\log n}} \to 0$ in distribution. But a sequence converging in distribution ...


1

For Polish $X$: Optimal measure $M^*$ exists: Topics in Optimal Transportation (Theorem 1.3 on page 19) and Optimal Transport (Theorem 4.1 on page 32) Equality $K_d=W_d$ holds: Topics in Optimal Transportation (Theorem 1.14 on page 34) and Optimal Transport (Remark 6.5 on page 95). The first of two books is more analysis-oriented, and may be easier to ...


1

I don't see how such a measure (call it $\mu$ for brevity) could exist. The definition of order implies that $I(1,\mathcal R)$ is precisely $C\setminus \{0\}$. Since $\lambda(C)=0$, we must have $\mu(C\setminus \{1\})=0$. Let $\alpha=\mu(\{1\}) = \mu(C)$. Next, for $s\notin C$ the set $I(s,\mathcal R)$ is the union of $C$ and the interval $[0,s)$. ...


1

Answer: It does not follow. I don't know what your lecturer meant or said, but you cannot draw this conclusion, as Michael's example shows. Let $(X_n)$ be a Markov chain with state space $\{0,1\}$ and transition matrix $P=\pmatrix{0&1\cr 1&0}$. The unique invariant distribution is $\pi=(1/2,1/2)$, but $\mathbb{P}_0(X_n=0)=(-1)^n$ does not converge ...



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