Explore Measurement Concepts with Our Examples and Questions

Consider $T:[0,1)\to <!--\; \to \; -->[0,1)$ to be aperiodic i.e $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->[0,1)$, $O(x)$ is infinite. $(O(x)=\{x,T(x),T^2(x),\cdots <!--\; \cdots \; -->\})$ and $T$ is not continuous. Imagine $\mathrm{\mu <!--\; \mu \; -->}$ is any $T$-invariant probability measure on $[0,1)$. It is written in my note that $\mathrm{\mu <!--\; \mu \; -->}$ is continuous because $T$ is aperiodic, but it is not kind of trivial to me. could anyone help me understand why the measure $\mathrm{\mu <!--\; \mu \; -->}$ is continuous?

Problem:
Let $\{X_i{\}}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ be a sequence of random variables on a probability space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},P)$ such that $\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{X}_i=X\text{ a.e.}$ a.e. Show that if $\underset{i}{sup}\text{E}(X_i^2)<\mathrm{\infty <!--\; \infty \; -->}$, then $\text{E}(X^2)<\mathrm{\infty <!--\; \infty \; -->}$.
My Attempt:
I will try to explain as best I can. First, I have a version of Fatou's Lemma stating that if $\{X_i{\}}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ is a sequence of non-negative random variables, then $\text{E}(\underset{i}{lim inf}{X}_i)\le <!--\; \le \; -->\underset{i}{lim inf}\text{E}(X_i)$.
If we let $Y_i=X_i^2$ then I have a sequence of non-negative random variables to work with. One concern of mine is this: can I assume that $\underset{i\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{X}_i^2=X^2$? I feel like that's necessary for for what I've written below to work.
If I can make that assumption then we have
$\begin{array}{rl}\text{E}(X^2)& =\text{E}(\underset{i}{lim inf}{X}_i^2)\mathbf{\text{ (Is this justified?)}}\\ & \le <!--\; \le \; -->\underset{i}{lim inf}\text{E}(X_i^2)\text{ (application of Fatou's Lemma)}\\ & \le <!--\; \le \; -->\underset{i}{sup}\text{E}(X_i^2)\text{ (property of real numbers)}\\ & <\mathrm{\infty <!--\; \infty \; -->}\text{ (by assumption)}.\end{array}$

Let $\mathbb{B}$ be a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on a set $X$ and let $f:X\to <!--\; \to \; -->\stackrel{¯<!--\; ¯\; -->}{R}$ be an extended real-valued function. Define the sets $P=f^{-<!--\; -\; -->1}(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->})$ and $N=f^{-<!--\; -\; -->1}(+\mathrm{\infty <!--\; \infty \; -->})$.
Define a real-valued function $g:X\to <!--\; \to \; -->\mathbb{R}$ by
$g(x)=\{\begin{array}{ll}f(x),& f(x)\in <!--\; \in \; -->\mathbb{R}\\ 0,& otherwise\end{array}.$
Prove that $f$ is $\mathbb{B}$-measurable if and only if $g$ is $\mathbb{B}$-measurable and $P,N\in <!--\; \in \; -->\mathbb{B}$.So i tried to prove this like this:
$P,N\in <!--\; \in \; -->\mathbb{B}$ obviously. Now assume $f$ is $\mathbb{B}$-measurable. Let $\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->\mathbb{R}$. Consider
$\{x\in <!--\; \in \; -->X|g(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}$=$\{x\in <!--\; \in \; -->X|f(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\setminus <!--\; \setminus \; -->P$. In this circumstance $P\in <!--\; \in \; -->\mathbb{B}$, $\{x\in <!--\; \in \; -->X|f(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\in <!--\; \in \; -->\mathbb{B}$ so $\{x\in <!--\; \in \; -->X|f(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\setminus <!--\; \setminus \; -->P\in <!--\; \in \; -->\mathbb{B}$. So $P^c\cap <!--\; \cap \; -->\{x\in <!--\; \in \; -->X|f(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\in <!--\; \in \; -->\mathbb{B}$. This is because, $\mathbb{B}$ is closed under finite intersection.
$(\Leftarrow <!--\; \Leftarrow \; -->)$$g$ is $\mathbb{B}$-measurable. Then
$\{x\in <!--\; \in \; -->X|f(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}=\{\begin{array}{ll}\{x\in <!--\; \in \; -->X|g(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\cup <!--\; \cup \; -->P,& \mathrm{\alpha <!--\; \alpha \; -->}>0\\ \{x\in <!--\; \in \; -->X|g(x)>\mathrm{\alpha <!--\; \alpha \; -->}\}\setminus <!--\; \setminus \; -->N,& \mathrm{\alpha <!--\; \alpha \; -->}\le <!--\; \le \; -->0\end{array}.$
Both of cases, observe that output sets belongs to $\mathbb{B}$.
But my professor said the proof was wrong. Where am I going wrong? Can you help me?

How to express the distribution of measurements ($X_m\sim <!--\; \sim \; -->{\mathcal{D}}_{\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}}$) given that the underlying physical process being measured follows a Gaussian distribution ($X_p\sim <!--\; \sim \; -->\mathcal{N}({\mathrm{\mu <!--\; \mu \; -->}}_p,{\mathrm{\sigma <!--\; \sigma \; -->}}_p)$) and the measurement uncertainty follows a Gaussian distribution ($\mathcal{N}({\mathrm{\mu <!--\; \mu \; -->}}_u,{\mathrm{\sigma <!--\; \sigma \; -->}}_u)$) too?
Note: ${\mathrm{\mu <!--\; \mu \; -->}}_u$ here is variable and takes the value of the specific instance/sample $X_p$.
Update:
Can one start in terms of PDFs as follows?
Given the PDF of the physical process:
$f_{X_p}(x)=\frac{1}{{\mathrm{\sigma <!--\; \sigma \; -->}}_p\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{1}{2}{\left(\frac{x-<!--\; -\; -->{\mathrm{\mu <!--\; \mu \; -->}}_p}{{\mathrm{\sigma <!--\; \sigma \; -->}}_p}\right)}^2}$
In order to express the PDF of the measurements affected by measurement error ($\mathcal{N}(X_p,{\mathrm{\sigma <!--\; \sigma \; -->}}_u)$), the idea is to use an specific sample t of the physical process as the mean of the measurement uncertainty distribution. Since the specific sample t follows a PDF $f_{X_p}(t)$, therefore integration is used over the whole range of t, and weighted by the PDF:
$f_{X_m}(x)={\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{1}{{\mathrm{\sigma <!--\; \sigma \; -->}}_u\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}}{e}^{-<!--\; -\; -->\frac{1}{2}{\left(\frac{x-<!--\; -\; -->t}{{\mathrm{\sigma <!--\; \sigma \; -->}}_u}\right)}^2}{f}_{X_p}(t)dt$
Is the above formulation reasonable?
Update after David K's answer: This question can also serve to get a more intuitive understanding of the convolution of two normal distributions.

I have 2 position estimates (along with their measurement error) and a difference in time between estimates. I estimate velocity using

Velocity = (PosA - PosB)/DeltaT

I am trying to estimate the error in my velocity estimate, but I can't seem to find any ways to calculate this. I assume it has to use Sigma_PosA and Sigma_PosB. I would also assume it's relative to DeltaT and/or abs(PosA - PosB). What is the velocity measurement variance/standard deviation?

I know that we can find the first weak derivative for $f(x)=|<!--\; |\; -->x|<!--\; |\; -->$, which is
$w(x)=\{\begin{array}{ll}-<!--\; -\; -->1& x<0\\ 0& x=0\\ 1& x>0\end{array}.$
But why is there no second weak derivative? If I try to find the first weak derivative of $w(x)$, I get that
${\int <!--\; \int \; -->}_{\mathbb{R}}v(x)\mathrm{\psi <!--\; \psi \; -->}(x)dx=-<!--\; -\; -->{\int <!--\; \int \; -->}_{\mathbb{R}}w(x){\mathrm{\psi <!--\; \psi \; -->}}^{\prime }(x)dx=2\mathrm{\psi <!--\; \psi \; -->}(0).$
What does this mean? Why is there no v such that $Dw=v$? I feel like I'm missing something and I'm not sure how to interpret this result.

Consider sequences of random variables $(X_n)$,$(Y_n)$ with $|X_n|<Y_n$. Suppose $(Y_n)$ converges in probability to $Y$, and $E{Y}_n\to <!--\; \to \; -->EY$. I have to show that, if $(X_n)$ converges to $X$ in probability, $E{X}_n\to <!--\; \to \; -->EX$. I have no idea of where to start. I have seen a question here in stackexchange that answers this for the case where $(Y_n)$ is constant, and would like some tips on how to do this. Can anyone help me?

My book says that ratio levels of measurement is the highest form of measurement and adheres to the same rules as as interval level measurement (distances between intervals of the scale are numerically equal), but it can have an absent property.
The examples are weight, height, blood pressure, pulse, etc.
How can weight or height be a ratio measurement? What thing being measured for weight has not weight? Or has no height? I understand no blood pressure or pulse.
Thanks :)

Here is an excercise of measure theory and real analysis. can you help me with it?
Let (X,M,$\mathrm{\mu <!--\; \mu \; -->}$) be a measure space and $f:X\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->}]$ a measurable function. Prove that
${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}>0\iff <!--\; \iff \; -->\mathrm{\mu <!--\; \mu \; -->}(\{x\in <!--\; \in \; -->X\mid <!--\; \mid \; -->f(x)>0\})>0.$
I tried to use the definition of the integral by the approximation theorem of measurable functions by simple functions, but I haven't been able to come to the conclusion.

Show that measuring the observable $|1⟩<!--\; ⟩\; -->⟨<!--\; ⟨\; -->1|$ is equivalent to measuring the observable Z up to a relabelling of the measurement outcomes.
What does it mean by relabelling of the measurement outcomes and how do I prove it? Could anyone help me?

I am trying to find an example that shows the image of a measurable function is not measurable.
Here is what I found: Let X={0,1} with $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra $\{\mathrm{\varnothing <!--\; \varnothing \; -->},X\}$. Let $f:X\to <!--\; \to \; -->X$ and f(x)=0. Then f(X)={0} is not measurable.
But how do we know that f is measurable in the first place? Obviously, we have $f^{-<!--\; -\; -->1}(\{\})=\{\}$. However, how can we say that $f^{-<!--\; -\; -->1}(X)=f^{-<!--\; -\; -->1}(\{0,1\})=X=\{0,1\}$? I mean f is not surjective; it is impossible that f(x)=1.

Assume that X is uniformly distributed on (0, 1) and that the conditional distribution of Y given X=x is a binomial distribution with parameters (n,x). Then we say that Y has a binomial distribution with fixed size n and random probability parameter.
I have to find the marginal distribution of Y. So I have to use that:
$P(Y=y)=\underset{x=y}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}P(Y=y|X=x)P(X=x)$
But when we have that X is uniformly distributed I think the sum becomes a integral so we get:
$P(Y=y)={\int <!--\; \int \; -->}_0^1(\genfrac{}{}{0ex}{}{n}{y})x^y(1-<!--\; -\; -->x{)}^{n-<!--\; -\; -->y}dx$
But what to do next? I think maybe take the binomial coefficient out from the integral, but how then using this to find the marginal distribution of Y? Hope anyone can help me? Do I also have to use that Y has a binomial distribution with fixed size n and random probability parameter?

In most measure theory text books one derives the Lebesgue anh Borel-Lebesgue measure from Caratheodory's extension to outer measures by first proving that the set of ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ measurable sets is a sigma algebra (the Lebesgue sigma algebra) and that ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ restricted to that sigma algebra is a measure, the Lebesgue measure. This measure space is even complete. However, then still one continues to show that ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ restricted to the sigmal algebra generated by the ring (on which the pre-measure was defined that was used to obtain ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ via Caratheodorys extension) is also a measure, the Borel-Lebesgue measure, and that the generated sigma algebra is the Borel sigma algebra.
So my question is, why is the Borel sigma-algebra "better" than the Lebesgue sigma algebra, because most of the time text books continue to work only on the Borel sigma algebra, even though the Lebesgue sigma algebra is its completetion and has some other favorable properties? I.e., I am just missing an argument in all the lecture notes and text books why we continue to work on the Borel sigma-algebra after having shown that the Lebesgue sigma algebra is larger (and after all we do all the extension from a pre-measure to an outer measure and then restricting to a measure because we want to get a larger set than just the ring on which we initially defined the pre-measure).

Given $X_i$ i.i.d random variables, $|X_i|\le <!--\; \le \; -->l$ for some $l>0$, $p_k=P(X_i=k)>0\mathrm{\forall <!--\; \forall \; -->}k$.
Let $N_n(k)=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{1}_{X_i=k}$ and let $\underset{n}{\prod <!--\; \prod \; -->}=\underset{k=1}{\overset{l}{\prod <!--\; \prod \; -->}}(p_k{)}^{(N_n(k))}$.
Prove the sequence $\frac{1}{n}\log <!--\; \; -->\underset{n}{\prod <!--\; \prod \; -->}$ converges almost surely and compute the limit.
I try this using the 3 Series theorem but I am unsure how to define the sequence to prove a.s. convergence using the 3 series theorem. Breaking down the sequence I arrive at the following:
$A_n=\frac{1}{n}\underset{k=1}{\overset{l}{\sum <!--\; \sum \; -->}}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{1}_{X_i=k}\log <!--\; \; -->(p(X_1=k))$
I'm not sure how to create the sequence though to prove the 3 criteria of the 3 series theorem if that is what is required for the proof.

We have the notion of Lebesgue measure, which I generally think of as the length/area of some interval in the space considered. We also have the notion of Lebesgue measurable functions. I was wondering if there is some way to compute the Lebesgue measure of a function? Perhaps, for example the function $f(x)=\sin <!--\; \; -->(x)$. The only thing that comes to my mind would be the arclength of the curve over some interval. Alternatively, can we describe the Lebesgue measure of a function as the length of the interval of the inverse image of the function?

Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be defined by
$f(x)=\{\begin{array}{ll}x[1+\sin <!--\; \; -->(\ln <!--\; \; -->x)]& \text{if }x>0\\ 0& \text{if }x=0\\ x+\sqrt{-<!--\; -\; -->x{\sin }^2<!--\; \; -->(\ln <!--\; \; -->|x|)}& \text{if }x<0.\end{array}$
Find the values of the four Dini derivatives of $f$ at $x=0.$
I could easily find out the values of the upper right $D^{+}f$ and lower right $D_{+}f$ Dini derivatives at $x=0,$ values being 2 and 0 respectively. But, in case of upper left Dini derivative of $f$ at $x=0,$ we have:
$\begin{array}{rl}{D}^{-<!--\; -\; -->}f(0)& =\underset{h\to <!--\; \to \; -->0-<!--\; -\; -->}{lim sup}\frac{f(h)-<!--\; -\; -->f(0)}{h}\\ \\ & =1+\underset{h\to <!--\; \to \; -->0-<!--\; -\; -->}{lim sup}\frac{\sqrt{-<!--\; -\; -->h{\sin }^2<!--\; \; -->(\ln <!--\; \; -->|h|)}}{h}.\end{array}$
Next putting: $-<!--\; -\; -->h=p$ so that $p\to <!--\; \to \; -->0+,$, I obtained:
$D^{-<!--\; -\; -->}f(0)=1-<!--\; -\; -->\underset{p\to <!--\; \to \; -->0+}{lim inf}\frac{|\sin <!--\; \; -->(\ln <!--\; \; -->|p|)|}{\sqrt{p}}.$
Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term $\sqrt{p}$ present in the denominator. I am confused. Please give some insights. Thanks in advance.

Let $f$ be a bounded probability density function. How do I show that ${\int <!--\; \int \; -->}_{\mathbb{R}}\mathrm{\sigma <!--\; \sigma \; -->}\exp <!--\; \; -->(-<!--\; -\; -->\frac{{\mathrm{\sigma <!--\; \sigma \; -->}}^2{\mathrm{\xi <!--\; \xi \; -->}}^2}{2})f(\mathrm{\xi <!--\; \xi \; -->})d\mathrm{\xi <!--\; \xi \; -->}\to <!--\; \to \; -->c<\mathrm{\infty <!--\; \infty \; -->}$ as $\mathrm{\sigma <!--\; \sigma \; -->}\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$?

I was thinking of doing dominated convergence theorem, but this doesn't work here. Could anyone give me any direction?

Suppose $X$ is a real random variable with some density function $f_X(x)$. Let $\mathbf{Y}=\mathbf{g}(X)$, where $\mathbf{g}(.):\mathbb{R}\to <!--\; \to \; -->{\mathbb{R}}^n$.
Is there a general way to represent the density $f_{\mathbf{Y}}(\mathbf{y})$?
For example, consider a simple case, $X\sim <!--\; \sim \; -->Uniform(0,1)$. $(Y_1,Y_2)=(X+1,X+2)$. Is there a closed form expression for the density of $(Y_1,Y_2)$ in ${\mathbb{R}}^2$?
Edit:
Dirac measures are permitted. In the above simple case, I presume the density could be
$f_{(Y_1,Y_2)}(y_1,y_2)=\{\begin{array}{lll}\mathrm{\delta <!--\; \delta \; -->}(y_1-<!--\; -\; -->y_2+1)& & 1\le <!--\; \le \; -->y_1\le <!--\; \le \; -->2\text{ and }2\le <!--\; \le \; -->y_2\le <!--\; \le \; -->3\\ 0& & \text{else}\end{array}$

Referring to the Doob's Maximal inequality we have that letting $M_t$ be a continous martingale, $\mathrm{\lambda <!--\; \lambda \; -->}>0$, $p>1$ and $T>0$ we have $P(\underset{0\le <!--\; \le \; -->t\le <!--\; \le \; -->T}{sup}|M_t|>\mathrm{\lambda <!--\; \lambda \; -->})\le <!--\; \le \; -->\frac{1}{{\mathrm{\lambda <!--\; \lambda \; -->}}^p}\mathbb{E}[|M_T{|}^p]$. I did not understand if the sup is the sup of the events ($|M_t|>\mathrm{\lambda <!--\; \lambda \; -->}$ : $t\in <!--\; \in \; -->[0,T]$) or if it is the sup of the random variable $|M_t|$ itself.

An equivalence relation is a binary relation that fulfills reflexivity, symmetry and transitivity.
A partially ordered set is a set equipped with a binary relation that fulfills reflexivity, antisymmetry and transitivity.
For a partially ordered set, some elements of the set might be incomparable, i.e. $x\le <!--\; \le \; -->y$ or $y\le <!--\; \le \; -->x$ may both not hold. That's why we use the term"partially". An example would be $\subset <!--\; \subset \; -->$ as a binary relation and the set M:={{1},{2},{1,2}}. Obviously, {1} and {2} are incomparable.
Question: Can also elements of a set, equipped with an equivalence relation, be incomparable?