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1

For every $x \in X$, we have three possibilities: $f(x) > 0$, so $f^+(x)$ is nonzero and $f^-(x)$ is zero, hence $x \in B$ and $x \not\in A$ $f(x) < 0$, so $f^-(x)$ is nonzero and $f^+(x)$ is zero, hence $x \in A$ and $x \not\in B$ $f(x) = 0$, so both $f^-(x)$ and $f^+(x)$ are zero, hence $x \in A$ and $x \in B$ In all three cases, $x$ is in either ...


1

Your set $S$ is a level set of a smooth function $f: \mathbb C^{n\times n }\to \mathbb C$, with $$ f(x,y)=x^Ty-x^Tx, $$ and hence is has measure zero. In fact $S$ can be viewed as a hypersurface in $\mathbb R^{2n\times 2n}$ of co-dimension $2$.


1

If you write out the equation in components and take the real and imaginary components of those constituent equations, you see that the set cuts out a proper real variety in $\mathbb{R}^{2n} \times \mathbb{R}^{2n}$, which thus has measure zero. Edit: In fact, you don't need to expand the equation at all: it's enough to see that (1) the equation is ...


1

Hint It is not generally true that $\sum P(A_{nm})=+\infty$. But it must be true for one of the $m$ following sequences: \begin{equation*} (B^{i}_{n})_{n\in\mathbb{N}}=(A_{nm+i})_{n\in\mathbb{N}}, \quad i=0,1..,m-1 \end{equation*}


2

For fixed $x \in \mathbb{R}$, define a mapping $$\tau_x: \mathbb{R} \to \mathbb{R}, y \mapsto y-x.$$ Then $\tau_x$ is continuous (hence measurable) and $\tau_x^{-1} = \tau_{-x}$. By definition, $$\int_{\mathbb{R}} |g(y-x)| \, m(dy) = \int_{\mathbb{R}} |g \circ \tau_x(y)| \, m(dy).$$ We can rewrite the right-hand side using image measures: $$ ...


1

First, let us be clear about the setup: We have an countably infinite number of coins; call them coin $1,2,3,\dots$. Now, for any finite set of coins (say coins $7,8,9,$ and $10$), we know the $\sigma$-algebra of events for those coin tosses (and we know the probabilities assigned). Now, to get a $\sigma$-algebra over the entire infinite sequence (and to ...


1

You just make sure you have countably many Lebesgue sets in your formulation, and from each of the disjoint sets remove sets of measure $0$ to make them Borel. Almost everywhere equality follows, and that's all the question cares about.


3

The notation $P(Y\mid X=x)$ probably means the distribution $\mu_x$ where the family $(\mu_x)_x$ is a conditioning of the distribution of $Y$ conditionally on $X$. Informally, $\mu_x(B)=P(Y\in B\mid X=x)$ for every Borel subset $B$ and every $x$. Formally, each function $G_B:x\mapsto\mu_x(B)$ is such that $P(Y\in B\mid X)=G_B(X)$ almost surely.


2

A non principal ultrafilter on the set of natural numbers is a tail set but it is far from Borel. What you are missing is that you need to make use of the fact that your set is Borel (or Lebesgue measurable) so that you can apply something like Lebesgue density theorem.


2

As commenters explained, the boundary of a compact set can have positive measure. How to proceed with the proof of $$\lim_{j \rightarrow \infty} \int_{K_j} f\, dm = \sup \left\{ \int_K\, f dm: K \subset \Omega, \ K \mbox{compact} \right\} $$ then? The inequality $\le$ holds because every $K_j$ is also eligible for the supremum on the right. To prove ...


0

Hint. Fact 1. If $f:\mathbb R\to [0,\infty]$ is Lebesgue measurable, then it expressed as $$ f(x)=\sum_{n\in\mathbb N}a_n\chi_{A_n}, $$ where $A_n$ Lebesgue measurable and $a_n\in [0,\infty]$ - This can be shown approximating $f$ from below by simple functions, and then using Lebesgue's Monotone Convergence Theorem. Fact 2. If $A$ is Lebesgue measurable, ...


0

Hint: A lebesgue measurable set in $\mathbb{R}$ is a Borel measurable set, upto sets of measure $0$.


3

By "real random variable" you mean the values are real? That is, we have $P(X=\infty)=0$ ? Then sets $$ A_n = \{ X < n \} $$ have finite measure, and $$ \Omega = \bigcup_{n} A_n . $$


1

It seems you have an idea of how to prove outer regularity (the analogous statement for open sets), so let us assume that for any $\epsilon > 0$ there is an open $U$ containing $E \in M$ such that $\mu(U\setminus E) < \epsilon$ ($\mu$ is Lebesgue measure). Step 1: Suppose $E$ is not closed (otherwise the statement is trivial), is bounded (will ...


0

See my comment above. The Haar measure assigns a notion of volume to a subset of a locally compact topological group. Let $\mu$ be a (left) Haar measure. Then there's an integral $\int_G f(x)d\mu(x)$, where $G$ is a locally compact topological group. Compare this to the definition of the integral of a real-valued function $f$ on a smooth manifold $M$: ...


2

In general $$ \eqalign{P(X + Y \le z) &= \iint_{\{(x,y): x+y \le z\}} dx\; dy\; f_{XY}(x,y)\cr &= \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy \; f_{XY}(x,y) }$$ If $X$ and $Y$ are independent, $f_{XY}(x,y) = f_X(x) f_Y(y)$ so this becomes $$ \int_{-\infty}^\infty dx \int_{-\infty}^{z-x} dy\;f_X(x) f_Y(y) = \int_{-\infty}^\infty dx\; ...


2

Consider $X_n\equiv 1+\log n$ everywhere, and $\lambda=1$. Then every element belongs to LHS but RHS is empty... so this is a counterexample.


3

Consider the map $$ \Phi : L^1 \to L^1, f \mapsto \varphi f. $$ By assumption, this is a well-defined linear map. Use the closed graph theorem to show that it is bounded (there are some details to fill in here). This shows that the functional $$ \psi : L^1 \to \Bbb{C}, f \mapsto \int f \varphi d\mu $$ is bounded (why)? Now use the characterisation of ...


0

The answer is min, provided by Michael in the comments. Thank you, Michael!


4

For $X=\mathbb R$, let, for each $n\in\mathbb N$, \begin{align*} f_n(x)=\begin{cases}1&\text{if $x\in(-\infty,0]$,}\\n^2&\text{if $x\in(0,1/n]$,}\\0&\text{if $x\in(1/n,\infty)$.}\end{cases} \end{align*} Let \begin{align*} f(x)=\begin{cases}1&\text{if $x\in(-\infty,0]$,}\\0&\text{if $x\in(0,\infty)$.}\end{cases} \end{align*} Consider ...


0

Continuity from above goes similar: $$\lim_n\mu(A_1\setminus A_{n+1})=\lim_n\sum_{k=1}^n\mu(A_k\setminus A_{k+1})=\mu(A_1\setminus A)$$ The remaining is $\mu(B\setminus A)=\mu(B)-\mu(A)$ for $A\subseteq B$.


0

If $A_n\downarrow A$, then $A_n^c\uparrow A^c$, then $\lim\mu(A_n^c)=\mu(A^c)$. As $\mu$ is finite, the measure of the universal set will be some finite $K$. Then we get $\lim(K-\mu(A_n))=K-\mu(A)$. Proved.


1

Uncountable additivity are considered especially in set theory. For example, an uncountable cardinal $\kappa$ is a real-valued measurable cardinal if and only if there exists a nontrivial $\kappa$-additive probability measure on $\kappa$. Although in some cases $\sigma$-additivity is enough. For example, the least cardinal which has a $\sigma$-additive ...


0

Mainly, sets and numbers rely on different footings. That brings structural similarities but also structural discrepancies. These allow one to study homomorphism as are measures only to a certain extend. To start with, both give rise to an algebraic structure by a binary operation admitting a neutral element: $$A\cup\varnothing=A=\varnothing\cup A$$ ...


0

First reduction: considering $f_k-f$ instead of $f$, we can assume that $f=0$. Second reduction: using an approximation of $g$ by simple functions and boundedness in $\mathbb L^p$ of $(f_k)$, we can reduce to the case where $g$ is the characteristic function of a set of finite measure, say $A$. Notice that $\int |f_k|\chi\{|f_k|>R\}d\mu\leqslant \lVert ...


0

$lim_{n \to \infty}||f_ng {||}_1=||fg {||}_1$ if and only if $lim_{n \to \infty}||f_ng - fg {||}_1=0$. Notice $||f_ng-fg{||}_1 \leq ||g{||}_{p'}||f_n-f{||}_{p}$. Hence you just need that $lim_{n \to \infty}||f_n-f{||}_{p}=0$. This amount to Lebesgue's Dominated Converge in $L^p$ spaces, which you have under the given assumptions. Let me know if you want a ...


0

Observe $|f| + |f_n| - |f-f_n|\geq 0$, with Fatou's lemma $$\liminf \int |f| + |f_n| - |f-f_n| \geq 2 \int |f| $$ by the assumption $\lim \int |f_n| = \int |f|$, the left-hand-side of above inequality also equals $$\liminf \int |f| + |f_n| - |f-f_n| = 2\int |f| -\limsup \int |f-f_n|,$$ combine them we get $$2\int |f| -\limsup \int |f-f_n|\geq 2 \int |f| $$ ...


1

No real answer but too much for a comment: $$\left\{ \text{number of heads infinite}\right\} =\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\left\{ \text{head at }k\text{-th toss}\right\} $$ showing that this event can be described by means of sets in $\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$. I am not sure, but suspect that ...


1

$g$ need not be continuous, but it has to be a Borel-measurable function $g: (\mathbb{R},\mathcal{B}(\mathbb{R})) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$. In order to conclude that $g(X)$ is $\sigma(X)/\mathcal{B}(\mathbb{R})$-measurable, note that $$(\Omega,\sigma(X)) \ni \omega \mapsto X(\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable ...


1

Beginner, I see your point... this passage: It turns out that once we specify the probability of every set that can be described in terms of finitely many coin tosses, the probability of every set in $\mathcal{F}_\infty$ is determined. is not immediately apparent. The point that he makes (I'm assuming) is that after this process has been repeated for ...


0

Note that $$\Omega = \bigcup_n \xi^{-1}([0,n])$$ It is easy to see that each set in the union has finite $\phi$ measure (because $P$ is a finite (probability) measure).


1

To prove the a.e. convergence, the method we think about may be the squeeze theorem. That is, we find two functions $F^*$ and $F_*$ that bounds the limit function. If the upper and lower bound are the same a.e., we can say that the limit exists a.e.. The natural way to choose the upper and lower bound, as we can see, is choosing $$F^*=\limsup F_n,F_* = ...


2

Let $(X_n)_{n=1}^\infty$ be a sequence of disjoint sets of strictly positive finite measure. Set $g_n = \mathbf{1}_{X_n}$. For each non-empty set $A\subset \mathbb{N}$ let $$f_A(x) = \sum_{n\in A}g_n(x).$$ Then $\|f_A - f_B\| = 1$ for distinct subsets $A,B\subseteq \mathbb{N}$. The power-set of $\mathbb{N}$ has cardinality continuum so $L_\infty(\mu)$ is ...


1

Try to adapt the following example: the functions $f_t(x) = \chi_{(-\infty,t]}(x)$ all belong to $L^\infty(\mathbb R)$, but satisfy $\|f_t - f_s\|_\infty = 1$ if $s \not= t$.


1

Simply connected sets can be nonmeasurable. Let $E\subset[0,1]$ be a nonmeasurable set and let $A=E\times[0,1]\cup[0,1]\times\{0\}$. If you draw a picture, you see that $A$ resembles a comb and is connected and simply connected. The set $A$ is not measurable. (This MSE question might be worth a look for the last fact.)


1

Hint: Consider a "fat cantor set" and use Rene's comment.


0

Let $\mu=\nu=m$, the Lebesgue measure, $g(y)=g(u,v)$. Then $\phi=\frac{d(\mu T^{-1})}{d\nu}=\frac{d(mT^{-1})}{dm}$. The function $\phi$ can also be seen roughly as $$\phi(y)=\lim_{y\in E,m(E)\rightarrow 0}\frac{mT^{-1}(E)}{m(E)}$$ for $y\in Y$. Obviously, it means the ratio before and after transformation for the aera of the set $E\subset Y$, i.e. the ...


2

First, we can show that $\overline{\mathcal B(\Bbb R)\times\mathcal B(\Bbb R)} = \mathcal B(\Bbb R^2)$, as every open disk -hence every open set in $\Bbb R^2$- can be covered by open rectangles (with sides parallel to the axes). Second, $f(x)$ can also be considered as a function in two variables (but constant in $y$). More precisely we consider ...


7

Put $$g_n(x) = \frac{x}{n} f(x) \chi_{[0,n]}(x)$$ Then $$\frac{1}{n}\int_0^n xf(x) dx = \int_{0}^\infty g_n(x) dx$$ Also, $|g_n(x)| \leq |f(x)| = f(x)$ for all $x$, and $g_n(x) \rightarrow 0$ pointwise. Therefore the dominated convergence theorem applies, and $$\begin{align} \lim_{n \rightarrow \infty} \frac{1}{n}\int_0^n xf(x) dx &= \lim_{n ...


3

The family of all subsets of $\mathbb{N}$ that are either finite or have a finite complement is an algebra, but not a $\sigma$-algebra.


3

We have $$\displaystyle \sum_{n=1}^\infty \mu\{x \in X : |f_n(x)| \geq \epsilon\} = \mu(\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}) < \infty$$ which means $\displaystyle \sum_{n=1}^\infty1_{ |f_n(x)| \geq \epsilon}$ is finite $\mu$-a.e. That is to say for any $\epsilon$, there are only finitely many $n$ such that $|f_n(x)| \geq \epsilon$ ...


1

Define random variable $Y$ by $\omega\mapsto 0$ if $X(\omega)<\infty$ and $\omega\mapsto +\infty$ otherwise. Then $0\leq Y\leq X$ hence $0\leq \mathbb E(Y)\leq\mathbb E(X)$. $Y$ only take the values in $\{0,+\infty\}$ and $P(Y=+\infty)=P(X=+\infty)>0$. This leads directly to $\mathbb E(Y)=(+\infty)\times P(Y=+\infty)=+\infty$


1

Since $X$ is positive, $E(X) \geq +\infty P(X = +\infty) = +\infty$. Otherwise, you can use the definition of $\int XdP$, it is defined as the sup of integration of simple functions that approaches $X$ from below. Remark integration of these simple functions on $\{X = +\infty\}$ tends to $+\infty$


3

No, such an $E$ need not be measurable: Let $A$ be a non-measurable set. Then $E=A\cup(\Bbb Q\setminus A)$ is non-measurable (otherwise, $A=\bigl( A\cup (\Bbb Q\setminus A)\bigr) \cap( \Bbb Q\setminus A)^C$ would be measurable). But, as $\Bbb Q\subset E$, every point $x$ in $E$ is a limit point of $E \setminus\{x\}$.


2

Lemma: Let $E \subset \Bbb{R}$ be uncountable. There is a countable set $F\subset E$ such that $E\setminus F$ satisfies your assumption. Proof: Let $$F := \{x \in E \mid \exists U_x \text{ neighbourhood of } x \text{ s.t. } U_x \cap E \text{ is countable}\}.$$ The open(!) covering $(U_x)_{x \in F}$ of $F$ has a countable subcover $(U_{x_n})_n$, (because ...


4

The idea is that, since the series $\sum\limits_nP(X_n\ne-1)$ converges, Borel-Cantelli lemma (the simple one) indicates that, almost surely, there exists some finite $N$ such that $X_n=-1$ for every $n\geqslant N$. Every such sequence $(X_n)$ is such that $S_n/n\to-1$ hence $P(S_n/n\to-1)=1$. The strong law of large numbers does not apply, although $(X_n)$ ...


4

First, $P(E_n)=1/n$ simply by symmetry, because $E_n$ happens when $\max\{X_1,X_2,\ldots,X_n\}$ is realized with $X_n$. Since the common distribution is continuous there is almost surely no ex aequo and $\max\{X_1,X_2,\ldots,X_n\}$ has as much chances to be realized with each $X_k$ with $1\leqslant k\leqslant n$, hence the result. Likewise, the ...


1

I don't really understand what you mean in facts 1 and 2, about $p$ being prime but may not be prime. But anyway: \begin{align*} P(H=n) &= \sum_{\substack{c,d\in\Bbb N \\ \gcd(c,d)=n}} P(X=c)P(Y=d) \\ &= \sum_{\substack{a,b\in\Bbb N \\ \gcd(a,b)=1}} P(X=na)P(Y=nb) \\ &= \sum_{\substack{a,b\in\Bbb N \\ \gcd(a,b)=1}} ...


0

From here: http://math.stackexchange.com/a/907506/140308 hint: $\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1 =\left \lfloor (n\log_2 (n+1)^2)+\log_2 (n+1)^2 \right \rfloor +1\ge \left \lfloor (n\log_2 (n+1)^2) \right \rfloor +\left \lfloor \log_2 (n+1)^2\right \rfloor+1$ now prove: $\left\lfloor (n\log_2 (n+1)^2) \right \rfloor \ge \left\lfloor ...


0

Sufficient condition As you said: if the inverse $S=T^{-1}$ exists and $S$ is measure preserving, then $$\mu(TE) = \mu(S^{-1} E )= \mu(E)$$ so $T$ is forward-measure-preserving. Necessary conditions For $T$ to be forward-measure-preserving, it must be "measure-wise injective", meaning that for every two measurable sets $A,B$, $$\mu(A\cap ...



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