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5

If $\mu (A)=\infty $ for all $A $, then $\mu $ doesn't distinguish between sets at all. You would have $\int f=\infty $ for all $f\geq0$ that are nonzero on at least one point, and you cannot even define the integral of any function that changes signs. So basically you get no measure theory whatsoever.


5

The theorem says that the graph of a real-valued function defined on $\mathbb{R}^n$ has measure zero in $\mathbb{R}^{n+1}$. E.g., if $f:\mathbb{R}\to \mathbb{R}$, then the curve defined by $f$ has measure $0$ in $\mathbb{R}^2$; if $f:\mathbb{R}^2\to \mathbb{R}$, then the surface defined by $f$ has measure $0$ in $\mathbb{R}^3$. The first problem you post is ...


3

Let $\mu$ be a finite measure on $(X,M)$, and let $\mu^*$ be the outer measure induced by $\mu$. Suppose that $E\subset X$ satisfies $\mu^*(E) = \mu^*(X)$ (but not that $E\in M$). a.) If $A,B\in M$ and $A\cap E = B\cap E$ then $\mu(A) = \mu(B)$ b.) Let $M_{E} = \{ A \cap E: A\in M\}$, and define the function $\nu$ on $M_E$ defined by $\nu(A\cap ...


3

Consider the function $$g(x):={\rm Av}_hf(x)={1\over h}\int_x^{x+h} f(t)\>dt\ .$$ Then $$g'(x)={1\over h}\bigl(f(x+h)-f(x)\bigr)={\rm Diff}_hf(x)\ .$$ It follows that $${\rm Av}_hf(b)-{\rm Av}_hf(a)=g(b)-g(a)=\int_a^b g'(x)\>dx=\int_a^b{\rm Diff}_hf(x)\>dx\ .$$


3

The forward implication should also be straightforward. Suppose $f(x)=g(x)$ for all $x \in \mathbb{N}$. Let $A \subseteq \mathbb{R}$ be the set on which $f$ and $g$ differ. What is the $\mu_\mathbb{N}$ measure of $A$?


3

For a counterexample with $n=1$, take $A$ and $B$ to be disjoint open subsets of $[0,1]$ defined as follows. Start with $C_0 = [0,1]$. At each stage $n \ge 1$, $C_{n-1}$ will be a finite union of closed intervals; take two disjoint finite sets of points $E_n$ and $F_n$ in the interior of $C_{n-1}$ such that every point of $C_n$ is within distance $1/n$ ...


3

The answer is negative. In fact if we define the Baire class $B_\alpha$ for countable ordinals $\alpha$ by saying $B_0=C(I)$, $B_{\alpha+1}$ is the set of pointwise limits of sequences in $B_\alpha$, and $B_\alpha=\bigcup_{\beta<\alpha}B_\beta$ for limit ordinals $\alpha$ then all the $B_\alpha$ are distinct. Or so I've read; don't ask me to prove it. ...


3

$f \equiv 1$ gives an example for which $\mu_f$ is complete, since it coincides with $\mu$. $f \equiv 0$ gives an example for which $\mu_f$ is not complete. Since $\mu_f(\mathbb{R})=0$, every set would have to be measurable, which we know is not the case.


3

Suppose there are measurable subsets $A,B$ and next to that some set $E$ with $A\subseteq E\subseteq B$. If $\mu(B-A)=0$ (and automatically $\mu A=\mu B$) then it is quite tempting to say that also $E$ can be labeled to be a set having measure $\mu A=\mu B$. That is the inspiration for "defining" $\mu^*(E)=\mu(A)=\mu(B)$. But wait a minute... What if ...


2

Let's break things down into steps: $L^p$ is a vector space. (Follows from their definition.) They are normed vector spaces: The $L^p$ norm, by definition, is a finite, nonnegative real number for given $f \in L^p$. 2.1 $\|f\|_p=0$ iff $f=0$ in $L^p$. This follows from the fact that if $f\neq 0$ on a set of positive measure, then $\int |f|^p >0.$ 2.2 ...


2

Let $A=\{x\in\mathbb{R}:f(x)\neq g(x)\}$. Then $f=g$ $\mu_\mathbb{N}$-a.e. if and only if $$ \mu_{\mathbb N}(A)=0 $$ if and only if $$ \#(A\cap\mathbb{N})=0. $$ Now, interpreting $ \#(A\cap\mathbb{N})=0 $ in words gives the proof.


2

This is false in general. For instance, you can take a Jordan curve $C$ in ${\mathbb R}^2$ of positive 2-dimensional measure (an "Osgood curve") and let $A, B$ be the components of ${\mathbb R}^2 -C$.


2

Let $\mathscr{C}$ be the collection of $E\in\mathcal{B}_X^*\otimes \mathcal{B}_Y$ for which: There exists $A\subseteq X$ null with $E\cup (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$. There exists $A\subseteq X$ null with $E\setminus (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$. Then it suffices to show that $\mathscr{C}$ is a $\sigma$-algebra ...


2

This is more or less the definition of uniform distribution. Properties we certainly expect from a uniform (on $[0,1]$) random variable $X$ are that we want $\Bbb P([0,1])=1$ and $\Bbb P([a,b])=\Bbb P([a+c,b+c])$ whenever $0\le a\le b\le b+c\le 1$. Together with additivity, this already leads to $\Bbb P(X\in E)=\mu(E\cap[0,1])$.


2

$(\Bbb R, \mathcal P(\Bbb R), \mu_\Bbb N)$ where $\mu_\Bbb N(A)= \vert {A \cap \Bbb N}\vert$= quantity of natural numbers in $A$, $\int f d\mu_\Bbb N$ =$\sum_{n=1}^{\infty} f(n)$. Define the space $L^1(\Bbb R, \mathcal P(\Bbb R),\mu_\Bbb N)=\{f:\Bbb R \to \Bbb R, \int_{\Bbb R} \vert f \vert d\mu_\Bbb N <\infty \}$ that is the space of the integrable ...


2

Let $m = \inf \limits_{x \in K} f(x)$ and $M = \sup \limits_{x \in K} f(x)$. Now $$m \, \lambda^d(K) \leq \int_K f \mathrm{d} \lambda^d \leq M\,\lambda^d(K).$$ Now, since $K$ is compact, there are $x_m, x_M \in K$ such that $f(x_m)=m$ and $f(x_M)=M$. SinceĀ $f$ is continuous and $K$ is connected, $f$ attains all values between $m$ and $M$. The claim ...


2

Let $F_n$ be the total fortune of gamblers $0,1,2,\ldots,n$ at time $n$. Then $G_n=F_n-n$ is a martingale with initial value $0$. And $F_{\tau_a}=2^4+2$. The $2^4$ is the fortune of gambler number $\tau_a-3$, and the $2$ is the fortune of gamble number $\tau_a$. All other gamblers are out of the picture (or have yet to start) at time $\tau_a$. By Doob's ...


2

Consider the standard coordiante system $(x,y)$ for $\mathbb{R}^2 \supset \Gamma(f)$. Let $P$ be the projection onto the $x$-axis, so that for any covering $C = \bigcup_{i} C_i$ of $\Gamma(f)$, we have that $P(C)=\bigcup_i P(C_i)$ is the projection of that covering onto the $x$-axis. Clearly, $[0,1] \subset P(C)$. Moreover, for any set $S \subset ...


2

Tonelli's theorem is not necessary since one of the sums is finite. Given two sequences $x_n$ and $y_n$ it is easy to prove using the definition of a limit that $\lim_n (x_n + y_n) = \lim_n x_n + \lim_n y_n$. This extends to the limit of any finite sum of sequences by induction. Hence, $$\sum_{i=1}^n\sum_{r=1}^\infty\alpha_i \mu(A_i \cap E_r) = ...


1

The problem is not to construct an infinite product of sigma-algebras. This always exists and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable. The problem is to find measures on this space. Of course a measure on the infinite product always pushes forward to a measure on the finite products contained in ...


1

You can take a set $N \times X_0$ where $N$ is Lebesgue measurable subset in $R^p$ with $\nu(N)=0$ and $X_0 \subseteq R^q$ is not Lebesgue measurable, i.e., $X_0 \notin L(R^q)$. Then set $N \times X_0$ has $\lambda$ measure zero because $\lambda$ is complete measure and $\lambda(N \times R^q)=0$. This set stands an example of a set which belongs to ...


1

After a lot of attempts I solved above problem: Let $s$ be a simple function such that $0\leqslant s\leqslant f$. Then $0\leqslant s\chi_E\leqslant f\chi_E$. Hence $\int \limits_{X}s\chi_Ed\mu\leqslant\int \limits_{X}f\chi_Ed\mu.$ But since $\int \limits_{X}s\chi_Ed\mu=\int \limits_{E}sd\mu$. Then $$\int \limits_{E}sd\mu\leqslant\int ...


1

For the remainder of the proof, just notice that you can approximate any non-negative, measurable function $f$ by an increasing sequence of simple functions $(s_n)_{n\in\mathbb N}$. Then, use monotone convergence for the integrals to conclude. Solution without monotone convergence for the integral: Notice that $$\int_E f = \sup\{\int_E s\colon s\le f\} = ...


1

If $S$ is any set, there is a measure $\mu$ on the set $X=\{0,1\}^S$ such that for each finite $F\subseteq S$ and each $a\in\{0,1\}^F$, $\mu(\{x\in X:x|_F=a\})=2^{-|F|}$ (this measure can be defined on the $\sigma$-algebra generated by these sets $\{x\in X:x|_F=a\}$ using the Caratheodory extension theorem, for instance). Intuitively, you can think of this ...


1

For part (a) you want to show that $A^c\cap B=B\setminus A$ has measure zero. Well note that $E\subset A\cup B^C$, this will give you the result you want by using your sets. Part (b) is pretty straightfowrward, it might be useful to prove additivity to look at $\overline A_i=A_i\setminus \cup_{j<i} A_i$ where $A_i\cap E$ is a disjoint family, since then ...


1

Perhaps there was some assumption about the simple functions in the proof of that particular result. In general it is not true, as the constant function $f = 1$ is a nonnegative simple function.


1

Exercise 20 - Let $\mu^*$ be an outer measure on $X$, $M^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, $\overline{\mu} = \mu^*|M^*$, and $\mu^+$ the outer measure induced by $\overline{\mu}$ as in (1.12) (with $\overline{\mu}$ and $M^*$ replacing $\mu_0$ and $\mathcal{A}$). a.) If $E\subset X$, we have $\mu^*(E)\leq \mu^+(E)$ with equality iff ...


1

No. Here is a counterexample: $g(x)=f(x+a)$ for $a\in\mathbb{R}$, $a\ne0$.


1

As Wikipedia points out, there are several results called Riesz (representation) theorem. A common feature of some of them is that on some spaces, every linear functional is integration against something. The result that the author uses here is the representation of linear functionals on $L^q$: they are integrals against $L^p$ functions. More precisely, ...


1

$$\lim_{t\to\infty}|f(x)|\chi_{\{|f|>t\}}=0 ~~~~\text{a.e.}$$ (Because $f\in L^1$ if we had a set of positive measure that was infinite, the integral would be infinite as well) And $|f(x)|\chi_{\{|f|>t\}}$ is dominated by an integrable function, namely $|f|$ (as $f\in L^1$), so we have: $$\lim_{t\to\infty} \int_{|f|>t}|f|~d\mu=\lim_{t\to\infty} ...



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