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17

If you insist on uncountable unions, then the measure of the unit interval is the sum of uncountably many measures of the individual points of the interval. These measures are either 0 (in which case $\mu(I) = 0$, which is bad) or nonzero, in which case $\mu(I)$ would be $\infty$,. A possibly more interesting question is whether finitely-additive measures ...


10

It's quite simple: Allowing only finite union yields too weak results [cf. Jordan measure] and allowing uncountable unions makes the system of measurable sets too big. If you for example assume that all singletons are measurable, then uncountable unions would imply that any set is measurable. This already doesn't work well with the Lebesgue measure, as the ...


6

Not necessarily. For example, consider the discrete topology on $\Bbb R,$ which has $$\mathcal B=\bigl\{\{x\}:x\in\Bbb R\bigr\}$$ as a base. The $\sigma$-algebra generated by $\mathcal B$ is the set of all subsets $A$ of $\Bbb R$ such that either (1) $A$ is finite or countably infinite, or (2) $\Bbb R\setminus A$ is finite or countably infinite. However, the ...


5

It is an unfortunate choice of terminology. The term measureable space, taken as a single unit, is defined to mean a set $X$ with a chosen sigma-algebra $\Sigma$. There is no specific choice of measure involved, and saying that $(X,\Sigma)$ is a measurable space is not intended to claim that it is possible to define a measure with that set and that ...


5

Another reason is that in a summable family $(x_i)_{i\in I}$, it can be shown the set $\;\{i\in I\mid x_i\neq 0\}$ is at most countable. Thus the theory of summable families comes down to the theory of series.


4

Countable has this very peculiar property of infinity. It can be indexed, such that every point is only a finite distance away. Namely, it can be indexed using $\Bbb N$. And $\Bbb N$ has that great property that given any $k\in\Bbb N$, you only need to travel finitely many steps to meet with $k$. Why is that important? Because (unfortunately) mathematics ...


3

For the answer to the first question: let $\mathcal S' = \{A \in \mathcal S : A^C \in \mathcal S\}$. See that $\mathcal S'$ contains all open sets, that $\mathcal S'$ is closed under taking complements, and that $\mathcal S'$ is closed under countable unions.


3

Observe that : $\left|\sin\left(\dfrac{x}{n}\right)\right| < \dfrac{x}{n} < x, x > 0$ and you can prove this fact. Thus it follows that : $|f_n(x)| < g(x) = \dfrac{x}{e^x-1}$. Next consider $f(x) = \sin\left(\dfrac{x}{n}\right) - \dfrac{x}{n} \to f'(x) = \dfrac{\cos\left(\dfrac{x}{n}\right)}{n} - \dfrac{1}{n} < 0\to f(x) < f(0) = 0 \to ...


3

the reason for your confusion could be, that you forgot some conditions for 3. Vitali sets are a counterexample that there is no measure on all the subsets of $\mathbb R$ with the following conditions: the measure of an open interval (a,b) should be b-a the measure should be invariant w.r.t to translation the measure is countable additive


3

Computing integral of Gaussian on whole line is possible using standard techniques. The problem is that no elementary antiderivative exists so you can't easily obtain integrals over arbitrary intervals. The function is integrable in the Lebesgue sense, just as it is integrable in "improper" Riemann sense. So Lebesgue theory doesn't introduce anything new ...


3

This is not correct. $f_n\to f$ a.e. iff $$\mu(\{x :\lim_{n\to\infty} |f_n(x)-f(x)|\ne0\}) = 0. $$ Convergence in measure has the limit outside the measure - one must take care when interchanging limit operations! Now, convergence in measure implies that there is a subsequence that converges a.e., and in a finite measure space (i.e. $\mu(X)<\infty$), ...


3

This is one of those things that is helpfully studied using an example. A very nice example for this issue is the "wandering block". Informally, the wandering block is the sequence of indicator functions of $[0,1],[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1]$, etc. More explicitly, it is the sequence $g_n(x)$ which comes about from enumerating the ...


3

First assume $f$ is nonnegative. Define $E=\bigcup_{n=1}^\infty E_n$, then as $f\chi_{E_n}\uparrow f\chi_E$ so by monotone convergence, \begin{align} \sum_{n=1}^\infty \int_{E_n} f\ \mathsf d\lambda &= \lim_{n\to\infty}\sum_{j=1}^n\int_{E_j} f\ \mathsf d\lambda \\ &=\lim_{n\to\infty} \int_{\bigcup_{j=1}^n E_j} f\ \mathsf d\lambda \\ ...


3

You are thinking too complicated. What we need is that the integral of the mollifier is $1$, and that the support of $\omega_\epsilon$ shrinks to the origin as $\varepsilon \to 0$ (if we took more general mollifiers, where the support need not be compact, we would still have $\lim\limits_{\epsilon \to 0} \int_{\lvert x\rvert > \delta} ...


2

This is not exactly an answer to the question as formulated, but it may clarify the precise formulation of the distinction. Convergence a.e. can be written using countable intersections and unions. First define $$A_{n,m} = \{ x : |f_n(x) - f(x)| > 1/m \}.$$ Now, for the convergence to fail at $x$, there must be some $m$ such that $x \in A_{n,m}$ for ...


2

Please note that, even if $A$ is finite, an element of $\prod _{\alpha \in A}M_\alpha$ is not a subset of $\prod _{\alpha \in A}X_\alpha$. So it can not be (or generate) a $\sigma$-algebra on $ \prod _{\alpha \in A}X_\alpha$. I think you meant to write: the $\sigma$-algebra generated by $\{\prod _{\alpha \in A} E_\alpha : E_\alpha \in M_\alpha, \alpha \in ...


2

Your proof so far is perfectly fine. All you need to do is prove the reverse inclusion and you'll be set.


2

Take a look further to the front of the book, maybe you can find the definition of the generated sigma-algebra that the author uses. Typically, the generated sigma-algebra is defined just in the way you described. In this case you only need to prove the second assertion of this exercise. However, there are also other definitions. One possible definition is ...


2

The "usual" Riemann integral is not defined, but the Lebesgue integral is and is equal to the "improper" Riemann integral: let $f(x) = x^{-\frac{1}{2}}$ on $(0,1]$ and $f(0)=0$. The $f$ is defined on $[0,1]$. Now define $f_n(x)=f(x)\cdot\chi _{(\frac{1}{n},1]}(x)$, Then, $f_n\nearrow f$ a.e. and so the Monotone Convergence Theorem applies to say that ...


2

Let's use a polar coordinate system with $M$ as the origin. Then the property can be stated as: the area of $M$ within the angular range $\alpha\le \theta\le \alpha+\pi$ is independent of $\alpha$. This area is $$\frac12 \int_\alpha^{\alpha+\pi} \int_0^\infty \chi_M(r,\theta) r\,dr\,d\theta$$ where $\chi_M$ is the characteristic function of $M$. ...


2

Let $n$ be such that $m^*(V)>1/n$, choose $n$ distinct rationals $q_1,\dots,q_n\in\mathbb{Q}/\mathbb{Z}$, and write $V_k=\bigcup_{i=1}^k V\oplus q_i$. Then if $V\oplus q_k$ is measurable, taking $A=V_k$ we find that $$m^*(V_k)=m^*(V\oplus q_k)+m^*(V_{k-1}).$$ Thus if $V$ were measurable, we could conclude by translation-invariance of $m^*$ and induction ...


2

A measure space is always a triple $(\Omega, \mathcal{F}, \mu)$, where the set $\Omega$ is the set with respect to which $\mathcal{F}$ is a sigma-algebra. The value of $\mu(\Omega)$ is called total mass of $\mu$. The measure $\mu$ is a probability measure iff its total mass is $1$. Now observe that $P(I \cap \Omega) = P(I) = P(I) \cdot P(\Omega)$, so both ...


2

An atom in a measurable space $(M,\mathcal{M})$ is a nonempty set $A\in\mathcal{M}$ such that $B\in\mathcal{M}$ and $B\subseteq A$ implies that either $B=\emptyset$ or $B=A$. If $\mathcal{M}$ is generated by a countable family of sets, so it is countably generated, every point in $M$ is in a unique atom and the atoms form a partition of $M$. A countably ...


2

These are just the pointwise max and min of $f$ and $g$, respectively. That is, $$(f\vee g)(x)=\max(f(x),g(x))$$ and $$(f\wedge g)(x)=\min(f(x),g(x)).$$ More generally, if $f$ and $g$ are elements of a poset, $f\vee g$ refers to the least upper bound of $f$ and $g$ and $f\wedge g$ refers to the greatest lower bound of $f$ and $g$ (if such elements exist). ...


2

Typically, we have $$(f\vee g)(x)=\max\bigl(f(x),g(x)\bigr)$$ and $$(f\wedge g)(x)=\min\bigl(f(x),g(x)\bigr).$$ This accords with a typical notion in boolean algebras that is often generalized to partially ordered sets, where $a\vee b$ and $a\wedge b$ indicate (respectively) the least upper bound and greatest lower bound of $a,b.$


2

Since $\{f_i, 1 \le i \le m\}$ is a set of Borel measurable functions $$ \{x:f_i(x)<c_i\}\subset \mathcal{B}(\mathbb R) $$ for all possible $c_i\in\Bbb{R}$. So $$ \{x:f_1(x)<c_1\}\times \{x:f_2(x)<c_2\}\times\cdots\times\{x:f_m(x)<c_m\}\subset \mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m) $$ for all possible $c_1\cdots c_m\in\Bbb{R}$. ...


2

If you are assuming $\mu (E)<\infty $, then I think this will work: Let $n\in \mathbb N$. Then using the definition of the outer measure, there is are open sets $U_{n+1}\subseteq U_n$ containing $E$ such that $\mu^* (E)=\mu(E)>\mu (U_n)-1/n$. Thus, $\tag1\mu (E)\geq \mu\left ( \bigcap _{n\in \mathbb N} U_n \right )$ (because $U_n\subseteq ...


2

The identity $$(a,b) = \bigcup_{n \in \mathbb{N}} [r_n(a),\infty) \cap \bigcup_{n \in \mathbb{N}} [r_n(b),\infty)$$ is not correct. As @uniquesolution pointed out in his (her) comment, you need to assume that $a<r_n(a)<r_n(b)<b$, but even then the identity does not hold true. Instead, it should read $$(a,b) = \bigcup_{n \in \mathbb{N}} ...


2

There is no such function, as you can see for $t \to 0$. But the point is that differentiability is a local property. Hence, it suffices for each interval $(t_0,t_1)$ with $0<t_0<t_1$ to find a dominating function which is independent of $t$ (but only needs to hold for $t$ in the above interval). This is easy to do, namely $x^2 e^{-t_0 x^2}$ will ...


2

Your $\chi_{D}u$ can be defined in that way: \begin{equation} (\chi_{D}u)(x)=\begin{cases} u(x), & x\in D, \\ \\ 0, & x\notin D. \end{cases} \end{equation} Edit: You don't need to mention that $u\in M\subseteq L^p[0,1]$ where $M$ is a bounded subset of $L^p[0,1]$. This is equivalent ...



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