Explore Measurement Concepts with Our Examples and Questions

Let $\mathcal{G}$ be any countable family of sets, where $\mathrm{\varphi <!--\; \varphi \; -->}\in <!--\; \in \; -->\mathcal{G}$.
Let ${\mathcal{G}}_1$ be the family of all finite union of difference of sets in $\mathcal{G}$.
Let ${\mathcal{G}}_2$ be the family of all finite union of difference of sets in ${\mathcal{G}}_1$.
May be I am misunderstanding the meaning of "finite union of difference" but to me here ${\mathcal{G}}_1={\mathcal{G}}_2$. However my text book says it is not necessary. How so?
This is used in the textbook to prove the ring generated by countable generator is countable.

As a mathematician I think of matrices as ${\mathbb{F}}^{m\times <!--\; \times \; -->n}$, where $\mathbb{F}$ is a field and usually $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. Units are not necessary.
However, some engineers have told me that they prefer matrices to have units such as metres, kg, pounds, dollars, etc. Assigning a unit of measurement to each entry to me seems restrictive (for instance if working with dollars then is $A^2$ allowed?).

Here are a few things that I would like to understand more deeply:
1. Are there examples where it is more appropriate to work with matrices that have units?
2. If units can only restrict the algebra, why should one assign units at all?
3. Is there anything exciting here, or is it just engineers trying to put physical interpretations on to matrix algebra?

This is a simple problem. Normaly I use two for-loops in GNU Octave to find the parameter $a$ and $b$ and $K$.
The cost function I'm going to minimize is:
$V=\underset{i=0}{\overset{N}{\sum <!--\; \sum \; -->}}(|u_i-<!--\; -\; -->K-<!--\; -\; -->u_0{a}^{-<!--\; -\; -->b{t}_i}|)$
Where $u_i$ is a vector for measurement and $t_i$ is also a measured vector. $u_0$ is known so only $a$,$b$,$K$ is unknow.
Is there any other way to minimize $V$ instead of using three for-loops in GNU Octave?
In my case, two-variables can solve the problem too. The $K$ is just a correction factor. Not necessary to 100%.

I've $F(t)={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{e^{-<!--\; -\; -->t{x}^3}}{1+x^4}dx$ and I have to see that it is well defined on the interval $(0,\mathrm{\infty <!--\; \infty \; -->})$.
For that, I have defined $f(x,t)=\frac{e^{-<!--\; -\; -->t{x}^3}}{1+x^4},x,t\in <!--\; \in \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$ so I have to see if $f$ is integrable in $(0,\mathrm{\infty <!--\; \infty \; -->})$.
We know that $f$ is integrable on ${\int <!--\; \int \; -->}_{(0,\mathrm{\infty <!--\; \infty \; -->})}|f|d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}$
$f(x,t)=|\frac{e^{-<!--\; -\; -->t{x}^3}}{1+x^4}|=\frac{e^{-<!--\; -\; -->t{x}^3}}{1+x^4}\le <!--\; \le \; -->e^{-<!--\; -\; -->t{x}^3}$
But hoe can I bound this? I have to bound it with an integrable function... but I don't know how to calculate the integral of $e^{-<!--\; -\; -->t{x}^3}$... Is there any other easier way to bound that? Or how can I solve my problem?

This seems to be a pretty basic question for this forum, please bear with me.
Consider that I have 10,000 measurements, say of the height of 10,000 people. These height measurements vary from 140 to 190 cm. I now define three height-groups: short (<150 cm), medium (150 to 170 cm) and tall (>170 cm). I can now calculate proportions of those groups in my set of 10,000 people (eg: “40% of the people are short, 30% are medium, 30% are tall”).
Now, consider that there is a random error associated with each of the height measurements. By a separate experiment, I concluded that this error distribution is well-approximated by a normal distribution, with a mean of zero and a standard deviation of 10 cm. That is, the measurers are unbiased, but do make some random errors.
Now, I would like to propagate this estimated error in height measurement to the proportions. That is, I would like to say something like “the percent of short men in $40\pm 3\mathrm{\%<!--\; \%\; -->}$” ($\pm$ could be standard error). Is there a theoretical way to go about this problem, rather than resorting to a Monte Carlo simulation?
The original data of 10,000 measurements could be described in two ways:
1. It is approximated by another normal distribution, of mean 165 cm and a standard deviation of 7.0 cm
2. It is described in a programming language data structure context; it is in a R vector, say "origData". Here, I am expecting R code that will take this vector and other inputs (from the question) and give me the standard errors.

There is a special non-commutative group related to the isometry $\mathrm{♯<!--\; ♯\; -->}:\mathbb{H}/\mathrm{♯<!--\; ♯\; -->}\to <!--\; \to \; -->{\mathcal{P}}_2\left({\mathbb{R}}^d\right)$, namely the set $\mathcal{G}(\mathrm{\Omega <!--\; \Omega \; -->})$ of Borel maps $S:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ (they lie in $\mathbb{H}$ ) that are almost everywhere invertible and have the same law as the identity map id.
Here
1/ $\mathrm{\Omega <!--\; \Omega \; -->}$ is the ball of unit volume in ${\mathbb{R}}^d$, centered at the origin.
2. $\mathbb{H}:=L^2(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{d}x,{\mathbb{R}}^d)$
My naïve guess is that "almost everywhere invertible" means the Lebesgue measure of $\{\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}\mid <!--\; \mid \; -->\mathrm{card}<!--\; \; -->(S^{-<!--\; -\; -->1}(\mathrm{\omega <!--\; \omega \; -->}))\le <!--\; \le \; -->1\}$ is 1.
Could you elaborate on this notion?

Let $(X,\mathcal{A})=([-<!--\; -\; -->1,1],{\mathcal{B}}_{\mathcal{[}\mathcal{-<!--\; -\; -->}\mathcal{1}\mathcal{,}\mathcal{1}\mathcal{]}})$ be a measurable space equipped with the Lebesgue measure, m.
Define $\mathrm{\varphi <!--\; \varphi \; -->}:{\mathcal{L}}^{\mathcal{1}}\mathcal{(}\mathcal{m}\mathcal{)}\to <!--\; \to \; -->\mathbb{R}$ by
$\mathrm{\varphi <!--\; \varphi \; -->}(f)=\int <!--\; \int \; -->(x-<!--\; -\; -->\frac{1}{2})f(x)dm(x)$
To find $||\mathrm{\varphi <!--\; \varphi \; -->}||$, I think the general strategy is to find an upper bound on $\mathrm{\varphi <!--\; \varphi \; -->}$, and then construct a function f such that it is attained.
In our case, we have
$||\mathrm{\varphi <!--\; \varphi \; -->}||=inf\{c\ge <!--\; \ge \; -->0:|\mathrm{\varphi <!--\; \varphi \; -->}(f)|\le <!--\; \le \; -->c||f||\}$. We have that $|\mathrm{\varphi <!--\; \varphi \; -->}(f)|\le <!--\; \le \; -->\int <!--\; \int \; -->|(x-<!--\; -\; -->\frac{1}{2})||f(x)|dm(x)$
I can split this up into two intervals: [−1,0] and [0,1]. On [0,1], we have that $|x-<!--\; -\; -->\frac{1}{2}|\le <!--\; \le \; -->\frac{1}{2}$. On [−1,0], we have that $|x-<!--\; -\; -->\frac{1}{2}|\le <!--\; \le \; -->\frac{3}{2}$. Hence $|\mathrm{\varphi <!--\; \varphi \; -->}(f)|\le <!--\; \le \; -->2\int <!--\; \int \; -->|f|$. This is my bound. However, I am unable to finish the proof because I am not sure how to construct a function f which attains this bound.

Hi once again a measurability question about Brownian motion. I'm not that familiar in how to prove that a function is measurable rigorously, so I'm stuck at the following exercise: Show that for $T>0$ the mapping
$\mathrm{\omega <!--\; \omega \; -->}\to <!--\; \to \; -->X_T(\mathrm{\omega <!--\; \omega \; -->})={\int <!--\; \int \; -->}_0^T{B}_t(\mathrm{\omega <!--\; \omega \; -->})dt$
is measurable. The Brownian motion is therefore assumed to be jointly measurable with continuous paths. How can I prove this? I know the definition of measurability that the preimage of measurable sets are measureable. But what are here the measurable sets in the image space. And how to proceed from there? Any help is really appreciated.

Given a Borel set $B$ of finite Lebesgue measure show that for every $\mathrm{ϵ<!--\; ϵ\; -->}>0$, there exist: a compact set $C_1\subseteq <!--\; \subseteq \; -->B$, a closed set $C_2\subseteq <!--\; \subseteq \; -->\mathbb{R}\setminus <!--\; \setminus \; -->B$ and a continuous function $\mathrm{\varphi <!--\; \varphi \; -->}:\mathbb{R}\to <!--\; \to \; -->[0,1]$ s.t.:
${\mathrm{\chi <!--\; \chi \; -->}}_{C_1}\le <!--\; \le \; -->\mathrm{\varphi <!--\; \varphi \; -->}\le <!--\; \le \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{\mathbb{R}\setminus <!--\; \setminus \; -->C_2}$
$||{\mathrm{\chi <!--\; \chi \; -->}}_B-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}|{|}_1<\mathrm{ϵ<!--\; ϵ\; -->}$"

Soo I have only the vague intuition of using the monotone convergence theorem somehow - e.g. perhaps setting $B\cap <!--\; \cap \; -->[n,n+1]$ as sets on which corresponding simple functions $f_n$ (such as the indicator function) can be defined so that they are wedged between ${\mathrm{\chi <!--\; \chi \; -->}}_{C_1}$ and ${\mathrm{\chi <!--\; \chi \; -->}}_{\mathbb{R}\setminus <!--\; \setminus \; -->C_2}$ (thus satisfying the first condition) and so that taking the limit on their integral would satisfy the second condition.
I am not sure how to go about accomplishing this, especially since the example I suggested for $B\cap <!--\; \cap \; -->[n,n+1]$ is not necessarily closed and therefore not necessarily compact. In addition, I am not sure how to define the required closed set $C_2$.
Any guidance on the steps of applying MCT (assuming I am at least right on this) to this problem would be much appreciated.
Edit: Another thought I had was perhaps using somehow the regularity of the Lebesgue measure to define the compact and closed sets (?), i.e. there are a compact set $K$ and an open set $U$ such that:
$\mathrm{\mu <!--\; \mu \; -->}(B)\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}(K)+\mathrm{ϵ<!--\; ϵ\; -->}$
and
$\mathrm{\mu <!--\; \mu \; -->}(B)\ge <!--\; \ge \; -->\mathrm{\mu <!--\; \mu \; -->}(U)-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$
Maybe these could be used to define the indicator function I mentioned above?

Show, that if $f_n\to <!--\; \to \; -->f$ and $f_n\to <!--\; \to \; -->g$ is $\mathrm{\mu <!--\; \mu \; -->}$-convergent, then $f=g$ almost everywhere on $X$
Hint
Use the fact, that:
$\{x\in <!--\; \in \; -->X:f(x)\ne <!--\; \ne \; -->g(x)\}=\underset{m=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}\{x\in <!--\; \in \; -->X:|f(x)-<!--\; -\; -->g(x)|\ge <!--\; \ge \; -->\frac{1}{m}\}$
So, I don't know how to use that hint. μ convergent means (correct me if I'm wrong), that
$f_n\to <!--\; \to \; -->f\text{ is }\mathrm{\mu <!--\; \mu \; -->}\text{ convergent}\iff <!--\; \iff \; -->\mathrm{\mu <!--\; \mu \; -->}(\{x\in <!--\; \in \; -->X:{\mathrm{\forall <!--\; \forall \; -->}}_{\mathrm{\epsilon <!--\; \epsilon \; -->}}\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}|f_n(x)-<!--\; -\; -->f(x)|>\mathrm{\epsilon <!--\; \epsilon \; -->}\})=0$
So I don't see it how the hint should be used.
It is not written what kind of measure our $\mathrm{\mu <!--\; \mu \; -->}$ is though, usually when there's nothing written we assume it's a Lebesgue measure, but I don't know if that has to be the case here

If the pre-image of the function is whole real line and is defined as following:
$f(x)=\{\begin{array}{ll}1& \text{if}x\in <!--\; \in \; -->\mathbb{Z}\\ 0& \text{otherwise}\end{array}$
What would be the essential supremum?
I understand that the essential supremum of a function is the smallest value that is larger or equal than the function values almost everywhere when allowing for ignoring what the function does at a set of points of measure zero.
Would it be still zero considering each individual integer essentially has measure zero? However, the measure of the integer set with value 1 is not zero, is it?
Thanks in advance.

If I have $m$ measurements to estimate an $n$ dimensional state vector, and I am using Kalman filter to do the filtering, then: Should I put all the $m$ measurements together in the measurement matrix (measurement transformation matrix ) and perform the filtering or, should I filter each measurement sequentially? Please provide some supporting explanation for your choice.
For e.g: Let $m=2$ and $n=3$. The state vector is 3 dimensional and we need to use the two measurements to get the posterior estimate the state vector. Now I can use one of these two methods:

1. Use all these measurements together to form a gain matrix of size $3\times <!--\; \times \; -->2$.
2. Use one measurement at one time and perform the filtering two times. The gain matrix will be $3\times <!--\; \times \; -->1$ in this case.

Which of the two methods is a better choice?

Invariance of domain theorem tells us that if a subset $V$ of ${\mathbb{R}}^n$ is homeomorphic to an open subset of ${\mathbb{R}}^n$, then $V$ must be open itself.
Question: If a subset $V$ of Rn is homeomorphic to a Borel subset of ${\mathbb{R}}^n$, must $V$ be Borel ?
Recall Borel(${\mathbb{R}}^n$) is defined to be the σ-algebra generated by the topology of ${\mathbb{R}}^n$.

Let $E$ be a normed $\mathbb{R}$-vector space, $(X_t{)}_{t\ge <!--\; \ge \; -->0}$ be an $E$-valued càdlàg Lévy process on a filtered probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},({\mathcal{F}}_t{)}_{t\ge <!--\; \ge \; -->0},\mathrm{P})$, $B\in <!--\; \in \; -->\mathcal{B}(E)$ with $0\notin \stackrel{¯<!--\; ¯\; -->}{B}$ and
$N_t(\mathrm{\omega <!--\; \omega \; -->}):=\left|\{s\in <!--\; \in \; -->(0,t]:\mathrm{\Delta <!--\; \Delta \; -->}{X}_s(\mathrm{\omega <!--\; \omega \; -->})\in <!--\; \in \; -->B\}\right|=\underset{\begin{array}{c}s\in <!--\; \in \; -->[0,t]\\ \mathrm{\Delta <!--\; \Delta \; -->}{X}_s(\mathrm{\omega <!--\; \omega \; -->})\end{array}}{\sum <!--\; \sum \; -->}{1}_B(\mathrm{\Delta <!--\; \Delta \; -->}{X}_s(\mathrm{\omega <!--\; \omega \; -->}))$
for $\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$ and $t\ge <!--\; \ge \; -->0$.

How do we see that $t\mapsto <!--\; \mapsto \; -->N_t(\mathrm{\omega <!--\; \omega \; -->})$ is càdlàg?

Suppose $(\mathbb{R},\mathrm{\tau <!--\; \tau \; -->})$ is the standard topological space. And $\mathbb{B}$ is the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra from this space.
Define set A as:
$A=\{x\in <!--\; \in \; -->\mathbb{R}:x=q_1\sqrt{n_1}+q_2\sqrt{n_2}\text{ for some }{q}_1,q_2\in <!--\; \in \; -->\mathbb{Q},\text{ and }{n}_1,n_2\in <!--\; \in \; -->\mathbb{N}\}$
How can I show that the set A belongs to $\mathbb{B}$?

Let X be a random variable with $E[X]=1$ and such that $X{1}_{\{X\le <!--\; \le \; -->0\}}=0$ a.s.
Question Does it follow that $X>0$ a.s.?
My progress I tried to write $X=X^{+}-<!--\; -\; -->X^{-<!--\; -\; -->}$, where both $X^{+},X^{-<!--\; -\; -->}$ are nonnegative, but I did not make any progress...

Let $A_1,A_2,\dots <!--\; \dots \; -->$ be a collection of events, and let $\mathcal{A}$ be the smallest $\mathrm{\sigma <!--\; \sigma \; -->}$-field of subsets of $\mathrm{\Omega <!--\; \Omega \; -->}$ which contains all of them. If $A\in <!--\; \in \; -->\mathcal{A}$, then there exists a sequence of events $\left\{C_n\right\}$ such that
$C_n\in <!--\; \in \; -->{\mathcal{A}}_n\text{ and }\mathbb{P}\left(A\mathrm{△<!--\; △\; -->}{C}_n\right)\to <!--\; \to \; -->0\text{ as }n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$
where ${\mathcal{A}}_n$ is the smallest $\mathrm{\sigma <!--\; \sigma \; -->}$-field which contains the finite collection $A_1,A_2,\dots <!--\; \dots \; -->,A_n.$
However, I do not know how to choose the $C_n$ such that they turn up in $\mathrm{\sigma <!--\; \sigma \; -->}(A_1,..A_n)$?

Someone can help me to verifying if the following reasoning it right?. Somebody read my proof and told me that there is a problem with the choose of the epsilon. If so, any one can clarify to me the mistake.

My proof is the following:

We can write
$A=\{x\in <!--\; \in \; -->X:\underset{n}{lim}{f}_n(x)\text{ exists}\}=\{x\in <!--\; \in \; -->X:\{f_n(x)\}\text{is a Cauchy sequence}\}.$
Let $\mathrm{ϵ<!--\; ϵ\; -->}>0.$ Define the sets $A_{m,n}(\mathrm{ϵ<!--\; ϵ\; -->})=\{x\in <!--\; \in \; -->X:|<!--\; |\; -->f_n(x)-<!--\; -\; -->f_m(x)|<!--\; |\; --><\mathrm{ϵ<!--\; ϵ\; -->}.\}$. This sets are measurable since the functions $f_n-<!--\; -\; -->f_m$ are measurable.

So, $x\in <!--\; \in \; -->A$ if and only there exists $N$ such that for $m,n\ge <!--\; \ge \; -->N$ $|<!--\; |\; -->f_n(x)-<!--\; -\; -->f_m(x)|<!--\; |\; --><\mathrm{ϵ<!--\; ϵ\; -->}$, that is to say, if only if
$x\in <!--\; \in \; -->\underset{m,n\ge <!--\; \ge \; -->N}{\bigcap <!--\; \bigcap \; -->}{A}_{m,n}(\mathrm{ϵ<!--\; ϵ\; -->}),$
and we have this if and only if
$x\in <!--\; \in \; -->\underset{p=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}\underset{m,n\ge <!--\; \ge \; -->p}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcap <!--\; \bigcap \; -->}}{A}_{m,n}(\mathrm{ϵ<!--\; ϵ\; -->}).$
Then we have that
$A=\underset{p=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}\underset{m,n\ge <!--\; \ge \; -->p}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcap <!--\; \bigcap \; -->}}{A}_{m,n}(\mathrm{ϵ<!--\; ϵ\; -->}).$
We conclude that $A$ is measurable.

So, let's say that I was given two independent random variables $\mathrm{\xi <!--\; \xi \; -->}$ and $\mathrm{\eta <!--\; \eta \; -->}$ and was told that $\mathrm{\xi <!--\; \xi \; -->}$ has continuous distribution (basically, $P(\mathrm{\xi <!--\; \xi \; -->}=c)=0\mathrm{\forall <!--\; \forall \; -->}c\in <!--\; \in \; -->\mathbb{R}$)
How can I prove that in such case their sum also will have continuous distribution? Problem that I have here is that I know that if two variables are independent then we can say $P_{\mathrm{\xi <!--\; \xi \; -->}+\mathrm{\eta <!--\; \eta \; -->}}=P_{\mathrm{\xi <!--\; \xi \; -->}}\ast <!--\; \ast \; -->P_{\mathrm{\eta <!--\; \eta \; -->}}$, where $\ast <!--\; \ast \; -->$ stands for measures convolution operation, but I am nor really sure how to compute such a convolution. Can you provide some explanation or intuition on how to calculate such integrals? In my case I need to know integral like this:
${\int <!--\; \int \; -->}_{{\mathbb{R}}^2}{\mathbb{1}}_{\{c\}}(x+y)d{P}_{\mathrm{\xi <!--\; \xi \; -->}}(x)d{P}_{\mathrm{\eta <!--\; \eta \; -->}}(y)$
And I have no rigorous way of showing that it is actually zero.

Is $f(x)=\sin <!--\; \; -->(\mathrm{sign}<!--\; \; -->x{e}^x)$ a measurable function ? I remember a theorem that if $f$ is a measurable function and $g$ is continuous then $g\circ <!--\; \circ \; -->f$ is measurable. Now we know that the sin function is continuous but how can I prove that $\mathrm{sign}<!--\; \; -->x{e}^x$ is measurable?