Explore Measurement Concepts with Our Examples and Questions

Suppose that every point of an interval $X=[0,1]$ lies in at least $k$ subintervals $A_i\subset <!--\; \subset \; -->X$, $i=1,...,n$. Show that
$k\le <!--\; \le \; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|A_i|$
where $|A_i|$ denotes the length of $A_i$.
My thoughts:
Let's consider $k$ copies of $X$. We will denote j-th copy as $X^j$.
If we would be able to cover them with $\{A_i{\}}_{i=1}^n$ then we're done.
(it is also enough to left uncovered the set of measure zero in every copy).
Let's fix an arbitrary point $x\in <!--\; \in \; -->X^1$. There exists $1\le <!--\; \le \; -->i_1<<!--\; <\; -->...<i_k\le <!--\; \le \; -->n$ such that $x\in <!--\; \in \; -->A_{i_j},j=1,...,k$. Then we will use $A_{i_j}$ to cover everything it can on the copy $X^j$. Now, let's choose another point $x\in <!--\; \in \; -->X^1$ and repeat the process.
There is a hard part in this proof. I can't understand why it can not be the case when we used all our $A_i$ but didn't fully cover every copy $X^j$(or didn't left uncovered the set of measure zero in every copy).

I have a measure $\mathrm{\mu <!--\; \mu \; -->}$ that is supported on $[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}$. What we are given is that, if we fix the first component $i$, then $\mathrm{\mu <!--\; \mu \; -->}(i,\cdot <!--\; \cdot \; -->)$ is a probability measure. Formally (maybe it should integrate to $\mathrm{d}i$ or similar, I cannot define this very well):
${\int <!--\; \int \; -->}_{x\in <!--\; \in \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)=1\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->[-<!--\; -\; -->3,3].$
I find it very hard to understand this tuple-indexing notation. My question is the following: What kind of assumptions do we need to have a result similar to:
${\int <!--\; \int \; -->}_{(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}\mathrm{d}i$

I just think that since for any fixed $i$ the measure $\mathrm{\mu <!--\; \mu \; -->}$ integrates to 1 (on the second dimension -- apologies for my poor terminology), then integrating over all $(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}$ should also give simply an 'iterated integral' where we first integrate wrt $x\in <!--\; \in \; -->\mathbb{R}$ for a fixed $i$, which will integrate to 1 , and then integrate over $i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]$. But of course, we cannot define
${\int <!--\; \int \; -->}_{(i,x)\in <!--\; \in \; -->[-<!--\; -\; -->3,3]\times <!--\; \times \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}{\int <!--\; \int \; -->}_{x\in <!--\; \in \; -->\mathbb{R}}\mathrm{d}\mathrm{\mu <!--\; \mu \; -->}(i,x)={\int <!--\; \int \; -->}_{i\in <!--\; \in \; -->[-<!--\; -\; -->3,3]}\mathrm{d}y$
where in the red parts I make an abuse of integartion rules.

How to calculate
${\int <!--\; \int \; -->}_Q{(x_1+x_2+\dots <!--\; \dots \; -->+x_n)}^{2}d{\mathrm{\lambda <!--\; \lambda \; -->}}_n$
whereas $n\ge <!--\; \ge \; -->2$ and
$Q=\{(x_1,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{R}}^n:0\le <!--\; \le \; -->x_i\le <!--\; \le \; -->1,i=1,\dots <!--\; \dots \; -->,n\}$

I know that I have to use induction and Fubini to calculate
$I(n)=\underset{0}{\overset{1}{\int <!--\; \int \; -->}}\underset{0}{\overset{1}{\int <!--\; \int \; -->}}\cdots <!--\; \cdots \; -->\underset{0}{\overset{1}{\int <!--\; \int \; -->}}{(x_1+x_2+\dots <!--\; \dots \; -->+x_n)}^{2}d{x}_n\cdots <!--\; \cdots \; -->d{x}_{2}d{x}_1$
And I already started to calculate the integrals for $n=2,3,4$ but the numbers I got did not enlighten me and I have no idea for an induction hypothesis. I also exchanged $x$ with 1 but I still have no clue. Maybe I am doing something wrong. Thanks for help in advance.

I am reading a proof concerning a conditional density $f(t|{\mathcal{G}}_{\mathcal{n}})$ where ${\mathcal{G}}_{\mathcal{n}}=\mathrm{\sigma <!--\; \sigma \; -->}(T_1,T_2,...,T_n)$ is a sub $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra generated by random variables $T_1,...,T_n$. In the proof, they express this conditional density in terms of the probability of $T_{n+1}$ being in some infinitesimal interval ds around t. They write:
$f(t|{\mathcal{G}}_{\mathcal{n}})=\frac{\mathbb{P}(T_{n+1}\in <!--\; \in \; -->[t,t+ds]|{\mathcal{G}}_n)}{ds}$
Intuitively, this makes sense to me, but I'm not really sure how to understand this more rigorously. I know that a conditioal probability given a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra is the same as a conditional expectation of an indicator function, but how do we make sense of this equality and the left hand side?
Edit: $f(t|{\mathcal{G}}_{\mathcal{n}})$ is the conditional density of event times in a point process given the first n points.

Let $X\sim <!--\; \sim \; -->\mathcal{N}(0,1)$ and define
$W=\{\begin{array}{ll}0& X<0\\ X& X\ge <!--\; \ge \; -->0\end{array}=X{\mathbf{1}}_{X\ge <!--\; \ge \; -->0}$
How can I calculate $\mathbb{E}\left[X|W\right]$ ?
Here's what I have tried so far:
I want to identify this random variable for any value w such that $W=w$. So I first noticed that
And now all I have to do is to calculate $\mathbb{E}[X|W=0]$. Since $W$ is neither a continuous or a discrete time variable, Im not sure how to express the conditioned density. Any help would be appreciated.
Thanks in advance.

Given that $\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_0(S_n^k)<{\mathrm{\mu <!--\; \mu \; -->}}^{\ast <!--\; \ast \; -->}(A)+\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{2^n}$, how do we justify the non-strict inequality $\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_0(S_n^k)\le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}^{\ast <!--\; \ast \; -->}(A_n)+\mathrm{ϵ<!--\; ϵ\; -->}$? Is there anything other to this than if $a<b$, then necessarily $a\le <!--\; \le \; -->b$?

I am not really sure if my answer is correct. Please enlighten me. Thank you.
The difference between the measures of an angle and its supplement is forty eight. Find the measure of both angles.
$$\begin{gathered} a+b=180 \\ b-<!--\; -\; -->a=48 \end{gathered}$$
$2a=132$ divided by 2 is $a=66$
$$\begin{gathered} 66+b=180 \\ b=180-<!--\; -\; -->66 \\ b=114 \end{gathered}$$
My answers: $a=66$ and $b=114$
Please correct me if I am wrong. Thank you.

I'm trying to understand what is the maximum deviation between two measurement methods. I read this on a paper on which there is a comparison between two devices to measure pH.
The sentence on the above paper is: "Maximum deviation was found to be 1.2% or 0.08 pH between two measurement methods.".

I am aware that using the counting measure on Lebesgue you can get important results. I am wondering if there is a way to exchange the limits using Dominated Convergence Theorem.
Let $a_{n,m}$ be indexed from 0 to infinity. $|a_{n,m}|<M$ for all indexes. $\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{n,m}$ exists for all m.
Does DCT allow you do say $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{n,m}=\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{n,m}$

Let $\mathrm{\gamma <!--\; \gamma \; -->}$ be a Gaussian measure on a locally convex space $X$ and let $q$ be a $\mathcal{E}(X{)}_{\mathrm{\gamma <!--\; \gamma \; -->}}$-measurable seminorm on $X$. Then the restriction of $q$ to the Cameron-Martín space $H(\mathrm{\gamma <!--\; \gamma \; -->})$ is continuous.
I have read the proof and it looks perfectly plausible, but the Cameron-Martin space has measure zero. Can't I just redefine q on $H(\mathrm{\gamma <!--\; \gamma \; -->})$ to be the absolute value of a discontinuous linear functional and get away with it?

Let $\nu$ denote counting measure and $\lambda$ denote Lebesgue measure. What is $(\nu \otimes \lambda )(\{(x,x)\}{)}_{x\in \mathbb{R}}$?
I am a little fuzzy on the product measure and have tried two ways to do this. Can someone explain why the incorrect version is wrong and the correct version is right?
By definition of the product measure $(\nu \otimes \lambda )(\{(x,x)\}{)}_{x\in \mathbb{R}}=\mathrm{\nu <!--\; \nu \; -->}(\{(x,x){\}}_{x\in <!--\; \in \; -->\mathbb{R}})\mathrm{\lambda <!--\; \lambda \; -->}(\{(x,x){\}}_{x\in <!--\; \in \; -->\mathbb{R}})=0$ since each $\mathrm{\lambda <!--\; \lambda \; -->}(\{(x,x)\})=0$ and
$(\nu \otimes \lambda )(\{(x,x)\}{)}_{x\in \mathbb{R}}=\int <!--\; \int \; -->\mathrm{\lambda <!--\; \lambda \; -->}(\{x\})({\mathrm{\chi <!--\; \chi \; -->}}_{\{(x,x)\}}{)}^{y}d\mathrm{\mu <!--\; \mu \; -->}=0$ and $(\nu \otimes \lambda )(\{(x,x)\}{)}_{x\in \mathbb{R}}=\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}(\{x\})({\mathrm{\chi <!--\; \chi \; -->}}_{\{(x,x)\}}{)}_{x}d\mathrm{\lambda <!--\; \lambda \; -->}=\mathrm{\infty <!--\; \infty \; -->}$. So this value is not well defined. If any of these are right, I would like to know why one is right, the other is wrong. If they are both wrong, I would appreciate an explanation on how to do this properly.

If a set $E\subset <!--\; \subset \; -->[a,b]$ has possitive measure, show that they exist $x,y\in <!--\; \in \; -->E$ s.t $x-<!--\; -\; -->y\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->\mathbb{Q}$
Because $m(E)>0$ then $\mathrm{\exists <!--\; \exists \; -->}\epsilon >0$ s.t $B(x,\epsilon )\subset <!--\; \subset \; -->E$ if not then $E=\{x_1,x_2,...\}$ and that can't be happening.
Now I can choose ${\epsilon }^{\prime }=\epsilon /\sqrt{5}$ such that $d(x,y)={\epsilon }^{\prime }$ and without loss of gennerality i can choose $x,y$ accordingly s.t $x-<!--\; -\; -->y>0$. Thus I get $d(x,y)=\epsilon /\sqrt{5}\Rightarrow <!--\; \Rightarrow \; -->x-<!--\; -\; -->y\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->\mathbb{Q}$
Is the solution ok ? How can I justify better that $B(x,\epsilon )\in <!--\; \in \; -->E$?

I am not that familiar in proving measurability of a function. The definition I know is that the preimage of measurable sets has to be measurable again. But how to prove that seems difficult to me at the first glance. I do have the following exercise: Let $(B_t)$ be a Brownian motion on $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},\mathbb{P})$ with continuous trajectories/paths. Consider the product space
$([0,\mathrm{\infty <!--\; \infty \; -->}),{\mathcal{B}}_{[0,\mathrm{\infty <!--\; \infty \; -->})},\mathrm{\lambda <!--\; \lambda \; -->})\otimes <!--\; \otimes \; -->(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},\mathbb{P})$
where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the Lebesgue measure. Show that the mapping $(t,\mathrm{\omega <!--\; \omega \; -->})\to <!--\; \to \; -->(B_t(\mathrm{\omega <!--\; \omega \; -->}))$($t\ge <!--\; \ge \; -->0$) defined from that space to $({\mathbb{R}}^{[0,\mathrm{\infty <!--\; \infty \; -->})},{\mathcal{B}}_{{\mathbb{R}}^{[0,\mathrm{\infty <!--\; \infty \; -->})}})$ is measurable. How can I prove this? I think the continuity is important, but I dont know how to involve it. Any help is appreciated.

I consider 2 linear Kalman filters (KF) that follow periodic cycles as shown here: KF(1) : Receive 1 measurement at time $k$, No measurements at time $k+1$, Receive 1 measurement at time $k+2$, No measurements at time $k+3$, etc.
KF(2) : Receive measurements at every time step.
In KF(1), whenever a measurement is missing, I simply apply time-update and ignore the measurement update step. Simulations show that in the steady-state, tr($P_{k|k}$) (the trace of the a-posteriori error covariance matrix computed after the reception of a measurement) is slightly greater in KF(1) than in KF(2), which intuitively makes sense. I would like to quantify this difference, or at least find upper and lower bounds for their ratio. Any leads or suggestions on where to look at or how to proceed?

Let $\{f_n\}$ be a sequence of measurable functions on a domain $E$ and p be a positive finite real number such that
(a) $\{f_n\}$ converges to a measurable function $f$ almost everywhere, and;
(b) $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(||f_n|{|}_p-<!--\; -\; -->||f|{|}_p)=0.$
Prove that $||f_n-<!--\; -\; -->f|{|}_p\to <!--\; \to \; -->0.$
I believe what we have to show is that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_E|f_n-<!--\; -\; -->f{|}^p=0.$
My approach is to hope to invoke the Dominated Convergence theorem, i.e., prove that there exists some $g\in <!--\; \in \; -->L^1(E)$ such that $|f_n|\le <!--\; \le \; -->g$ for almost everywhere for all $n$, then we can have
${\int <!--\; \int \; -->}_E|f_n-<!--\; -\; -->f|\to <!--\; \to \; -->0.$
Then by (a), we can find an integer $K$ such that
$|f_k-<!--\; -\; -->f|<1$
for all $k\ge <!--\; \ge \; -->K$. Then
$0\le <!--\; \le \; -->{\int <!--\; \int \; -->}_E|f_k-<!--\; -\; -->f{|}^p\le <!--\; \le \; -->{\int <!--\; \int \; -->}_E|f_k-<!--\; -\; -->f|.$
Since the integral on the right approaches 0, we conclude that the same is true for the middle term and we will be done.
However, I have no idea how to obtain the $g$ as mentioned before, as well as use condition (b) given in the question. Some help would be appreciated.

Ok, so this is a bit of an odd question. It is sort of a physics question, but I felt it was more appropriate here. Feel free to suggest another venue.
Let's just look at a simple example. We'll use kilograms as an example unit of measurement. Let's say we have three objects, A, B, and C with masses 11 kg, 12 kg, and 23 kg. It is clear there is an additive relationship here that has a real physical meaning. The third object really does have the equivalent amount of matter as the first two combined. This real relationship is played out with how physical laws play out. We can go down the philosophical rabbit hole here, but I don't think it is productive.
A measurement in a given unit must be equivalent to that number times the unit, i.e. $11kg=11\times <!--\; \times \; -->(1kg)$. Of course, multiplication can be defined in terms of addition. So we actually require that
$11kg=\underset{\text{11 terms}}{\underset{⏟<!--\; ⏟\; -->}{1kg+1kg+\cdots <!--\; \cdots \; -->1kg}}.$
In a real physical sense, the amount of stuff in eleven objects each with 1 kg mass really is equivalent to the amount of stuff in one object with 11 kg of mass.
So here is the main question: Is number of kilograms beyond 10 kilograms a valid unit of measurement? Let us denote it $k{g}_{10}$
Now our objects have "masses" $1k{g}_{10}$, $2k{g}_{10}$, and $13k{g}_{10}$. These aren't actually masses, they are "masses beyond 10kg". The additive relationship from our original unit of kg does not carry over to our new "unit" of $k{g}_{10}$.
Can we even say that $1k{g}_{10}$ + $1k{g}_{10}$ = $2k{g}_{10}$? What are we actually adding here? We are not adding kilograms! If I have an object that has mass 11 kg ($1k{g}_{10}$) and I increase its mass by 1 kg, it now has mass 12 kg ($2k{g}_{10}$) so it would seem that $1k{g}_{10}$ + 1 kg = $2k{g}_{10}$.
So, maybe this is a stupid thing to worry about, but it just seems that we should have rigorous set of rules about what constitutes a valid unit of measurement. Those rules would include something about additivity and the relation to the thing being measured and not being able to just arbitrarily set zero anywhere.
I would argue that $k{g}_{10}$ is not technically a valid unit of measurement, and that this is reasonable since the unit of measurement is actually kilograms, which is a true unit of measurement (arithmetic operations on it make sense in terms of the physical thing it is actually measuring), but for whatever reason, we might be interested only in kilograms beyond 10.

I am a newby to Kalmar filters, but after some study, I think I understand how it works now. For my application, I need a Kalmar filter that combines the measurement input from two sources. In the standard Kalmar filter, that is no problem at all, but it assumes that the measurement inputs from the two sensors are available at the same times. In my application, there is one new measurement from sensor 'b' for every 13 measurements of sensor 'a'.That is, 12 out of 13 times, the measurement of sensor 'b' is missing.
How would you handle that normally? Do you simply use the predicted measurements values as substitute for the missing ones? Does that not lead to overconfidence in the missing measurements? How else can it be handled?

Revised based on comments.
Measure-theoretic probability is used in finance (among other fields).
However, was wondering where unbounded measures are used, outside of mathematics and physics. For example, is there a use for unbounded measures in finance or biology or business?
Apologies for a broad question -- there really is not 'right answer' but hoping this group will have suggestions. Many thanks!

Let $n\in <!--\; \in \; -->N$, $(X,A,\mathrm{\mu <!--\; \mu \; -->})$ be a measure space and $E_1,...,E_n\in <!--\; \in \; -->A$. For each $j\in <!--\; \in \; -->\{1,...,n\}$ define
$C_j:=\{x\in <!--\; \in \; -->X|x\in <!--\; \in \; -->E_k\text{for exactly }j\text{ indices},k\in \{1...,n\}\}.$
I want to find a general expression for all the $C_j$ since I need to prove that each $C_j$ is a measurable set, I made an example and it is as follows
With $n=3$
$C_1=(E_1\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)$
$C_2=(E_1\cap <!--\; \cap \; -->E_2\cap <!--\; \cap \; -->E_3^c)\cup <!--\; \cup \; -->(E_1\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)\cup <!--\; \cup \; -->(E_1^c\cap <!--\; \cap \; -->E_2^c\cap <!--\; \cap \; -->E_3)$
$C_3=\underset{k=1}{\overset{3}{\bigcap <!--\; \bigcap \; -->}}{E}_k$
So for $C_n$ we already have a general expression:
$C_n=\underset{k=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{E}_k$
for $C_1$:
$C_1=\underset{j=1}{\overset{n}{\bigcup <!--\; \bigcup \; -->}}(E_j\cap <!--\; \cap \; -->\underset{k\ne <!--\; \ne \; -->j}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{E}_k^c)$
I was thinking of introducing permutations but actually I don't know if it works, someone can guide me a bit to express the $C_j$ sets in a manageable way

What are the measurement units of $\mathrm{\omega <!--\; \omega \; -->}$ in this equation and why
$Y(x,t)=A\cos <!--\; \; -->(kx-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}t)?$