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19

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...


10

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...


8

Suppose for a contradiction that $E+a$ always contains a rational. Then $$ \bigcup_{q\in \mathbb Q}(E-q) $$ would contain every $a\in \mathbb R$, but would also be a countable union of measure-zero sets.


7

Take on $\mathbb R$ the measure $d\mu = \frac{dx}{1+x^2}$. Then for every non constant convex function $f(x)$ if the integral is well defined one has $$ \int f(x) d\mu(x) = +\infty $$ since $f(x)>mx$ for either $x\to +\infty$ or $x\to-\infty$. For constant functions the integral only depends on the total mass of the measure $\mu$. Hence you cannot ...


5

The one dimensional case The result is true if we assume that some integrals are finite. Below I will assume that $\int_0^\infty x \,\mu(dx)=\int_0^\infty x \,\nu(dx)<\infty.$ Emanuele Paolini's argument shows that some such assumption is necessary. Fix $z\in\mathbb{R}$ and $m>0$. The functions $f(x)=(m(x-z)+1)_+$ and $g(x)=m(x-z)_+$ are both ...


5

The solution method for these kinds of problems is almost always to expand the troublesome part (here $\frac{1}{e^t-x}$) of the integrand in some series and integrating term by term. In some cases (usually if the integrand has a periodic factor) one can split the integration over $[0,\infty)$ into pieces (e.g. $[0,2\pi]$,$[2\pi,4\pi]$,$\ldots$) and then ...


4

A continuous map from $\Bbb R^2$ to $\Bbb R$ cannot be injective. This is trivial; proof below. It's higher-dimensional versions, like the fact that a continuous map from $\Bbb R^3$ to $\Bbb R^2$ cannot be injective, that I suspect are non-trivial. Given two points $a,b\in\Bbb R^2$, let $[a,b]$ denote the line segment from $a$ to $b$. Say $f:\Bbb ...


4

Consider the sequence defined by $$f_n = \begin{cases} \chi_{[0,1]} & n = 1\\ \chi_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} & 2^k \le n < 2^{k+1} \end{cases}$$ Then $\|f_n\|_1 = \frac1{2^k} \to 0$ but $f_n \not\to 1$ on any $x\in[0,1]$. You can, however, prove that there exists an a.e. convergent subsequence with limit $0$ in this ...


3

The weak topology has the property that the dual is the same as the dual with respect to the original topology, if $(E,\tau)$ is a topological vector space, and $E^\ast$ its (topological) dual, then $$(E,\sigma(E,E^\ast))^\ast = E^\ast.$$ For the topology of pointwise convergence on $C(S)$, the dual is easily described: if $\lambda \colon C(S) \to ...


3

$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e. $$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$ for any Borel set $B$. This implies $$\int g(x) \, \nu_1(dx) = \int g(f(x)) \, \nu(dx)$$ for any $g \in L^1(\nu_1)$.


3

Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation. Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on $(M, ...


3

Let's check whether $\mu_A$ satisfies the conditions of being a Radon measure step by step. 1. $\mu_A$ is a measure. This is easy, I leave it to you. 2. $\mu_A$ is finite on compact sets. Let $K\subseteq X$ be compact. Then, $$\mu_A(K)=\mu(K\cap A)\leq \mu(K)<\infty,$$ given that $\mu$ is a Radon measure. 3. $\mu_A$ is outer regular on all Borel sets. ...


3

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$ \{ x: \in \mathbb R : f(x) > a \}, $$ although it is not always measurable.


3

Let $\Omega = \{a,b\}$. Define $X,Y:\Omega\to\mathbb R$ by $$X(\omega) = \begin{cases}0,& \omega=a\\ 1,&\omega=b. \end{cases} $$ and $$Y(\omega) = \begin{cases}1,& \omega=a\\ 0,&\omega=b. \end{cases} $$ Then $\sigma(X)=\sigma(Y) = 2^\Omega$, but $XY$ is identically zero so $\sigma(XY)=\{\varnothing,\Omega\}$.


3

Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$. This sort of thing is usually how filtrations are created.


3

Let $X$ be a non-empty set and $\mathscr A$ an algebra on it. A premeasure on a $\mathscr A$ is a function $\lambda:\mathscr A\to[0,\infty]$ such that $\lambda(\varnothing)=0$; and if $A_1,A_2,\ldots$ is a countable collection of disjoint sets in $\mathscr A$ and if their union is contained in $\mathscr A$, then $$\lambda\left(\bigcup_{n=1}^{\infty} ...


3

Hints: enumerate any countable subset of $\mathbb R$ as, say, $\{a_n\}_n \ge 1$. Now cover $a_n$ by an interval of length $\frac{\varepsilon}{2^n}$. What is the total measure of the cover? Notice that $\varepsilon>0$ may be chosen arbitrarily. What does that imply?


2

Question 19 For a complex measure $\gamma$ and a set $B$, notice that $\gamma(A) = 0$ for every measurable $A \subset B$ iff $\lvert\gamma\rvert(B) = 0$. And this happens iff $\lvert\gamma\rvert(A) = 0$ for every measurable $A \subset B$. From the above, $\nu \perp \mu$ iff $\lvert\nu\rvert \perp \lvert\mu\rvert$. And, \begin{align*} \lvert\nu\rvert \ll ...


2

Your constant $N_1$ depends on $k$. Here is a simplified argument that corrects the error. The final inquality will be helpful. Let $F_j = \{x : f(x) > 1/j\}$. Then $F_1 \subset F_2 \subset F_3 \subset \cdots$ and $m(\cup F_j) = 1$ since $f > 0$ almost everywhere. Apply continuity from below to see $m(F_j) \to 1$. Fix $\epsilon > 0$. Since ...


2

Suppose there were such continuous maps. Then $\eta$ would be injective. Hence it's restriction to the closed unit disc is a homeomorphism. The image of that disc would be a closed bounded interval. But the disc is not homeomorphic to an interval: removing an interior point from the interval leaves a disconnected set, while removing any point from the disc ...


2

Yes it can be uncountable, let ${A_n}=\{n\}$, so the sets are $\{1\},\{2\},\{3\}\dots$ Then it is clear that the sets that are the union of two or more of these sets are the subsets of $\mathbb N$ with at least two elements. This subset is not countable, we can find a bijection with the set of infinite binary sequences with at least two ones (which is ...


2

By the (multi-dimensional) Leibniz formula, we have $$\begin{align*} \partial^{\alpha} \phi_k&= \sum_{\beta+\gamma = \alpha} \underbrace{c_{\beta,\gamma} (\partial^{\beta} \phi) \cdot (\partial^{\gamma} (1-\psi(k \bullet)))}_{=:S_{\beta,\gamma}} \end{align*}$$ for some constants $c_{\beta,\gamma}$ (which can be calculated expliticly). If $\gamma=0$, ...


2

a) I leave it to you to show that $\phi \chi_U$ is lower semi continuous. The hint they gave you is actually a part of Proposition 7.14. Corollary 7.13 actually reads $$ \int f \, d\mu = \sup \{\int g \,d\mu \,\mid\, 0\leq g \leq f \text{and } g\in C_c\} $$ if $f$ is lower semicontinuous. With this information, first try part (a) again. If you don's ...


2

From wikipedia: $\:$ "In particular Krivine (1969) showed there was a model of ZFC in which every ordinal-definable set of reals is measurable, ..." Therefore ZFC does not prove that there is a definable nonmeasurable set. Using an explicit definable well-order on $\mathbb{R}\hspace{-0.04 in}\cap$OD to run a Vitali construction gives an explicit set ...


2

If you can prove that your definition defines something, then will in particular define something in Solovay's model, and that something will in particular not be a nonmeasurable set. So the best you can hope for is an explicit definition of a subset of $\mathbb R$, such that the thing defined will in some models of ZF be a nonmeasurable set. And this is ...


2

From $f_n - f\to 0$ in $L^p$ follows that there exists a subsequence of $f_n$ which converges almost everywhere to $f$. Thus, $f\ge 1$ almost everywhere. Now, notice that $$ \left| \frac 1 {f_n} - \frac1f \right | = \frac{|f_n - f|}{f_n f} \le |f_n - f|. $$ Integrate (take essential supremum) on both side to obtain the statement.


2

Assume $\lambda(A)>0$ (otherwise just take $K=\emptyset$). By the inner regularity of Lebesgue measure, there is a compact set $K_0 \subseteq A$ with $\lambda(K_0)>\lambda(A)/2$. Define $$g(x)=\lambda(K_0 \cap [-x,x])$$ By the Intermediate Value Theorem, there exists an $x$ such that $g(x)=\lambda(A)/2$, and then $K:=K_0 \cap [-x,x]$ will be a compact ...


2

Your set looks countable to me. If I parsed it correctly, the hint is that any family of pairwire disjoint sets of positive measure is countable.


2

Assuming that $C \subset C_1$ - the exercise makes not much sense and the assertion is in general false without that assumption - we have \begin{align} \mu(C_1) &= \mu\bigl((C_1 \cap A_k) \cup (C_1 \setminus A_k)\bigr)\\ &= \mu(C_1 \cap A_k) + \mu(C_1\setminus A_k)\\ &\geqslant \mu^\ast(C\cap A_k) + \mu^\ast(C\setminus A_k)\\ &\geqslant ...


2

The usual definition to make in this context is to say that the "valueless" part is the largest open set which has zero value. As you mention, there is typically no largest set which has zero value. However, there is always a largest open set of zero value: take the union of all open sets of zero value. This union will still have zero value, because it is ...



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