Explore Measurement Concepts with Our Examples and Questions

Reading this document, when measuring a thing we have
$\text{relative error}=\frac{\text{absolute error}}{\text{value of thing measured}}.$
Let's say we have a measurement of the height of an object as 45.2m with error of $\pm <!--\; \pm \; -->0.05m$.
At that point our relative error would be $\frac{0.05m}{45.2m}$.
However, if a person who measures heights always rounds his measurement to three significant digits would it change the relative error?
I am not sure if this is correct, but in other words would the answer 45.2m+0.05m=45.25m rounded to 45.3m as the approximate value. After that would th e absolute error be 0.1m, so that the relative error would be $\frac{0.1}{45.2}$?

I am wondering if there is a basic unit in math.
$3cm\times <!--\; \times \; -->2cm=6c{m}^2$
The cm is a unit of measurement, but what about:
$3cm\times <!--\; \times \; -->2=6cm$
Should the 2 have a unit of counting? Like I have 2 (amount of stuff) 3 cm strings.
Should it be something like:
$3cm\times <!--\; \times \; -->2m{u}^0=6cm$
(where mu is a "mathematical" unit)
The unit is to the power of 0 because if it was not it would do this:
$3mu\times <!--\; \times \; -->2mu=6m{u}^2$
And I don't think this works right, but $m{u}^0$ seems to:
$3m{u}^0\times <!--\; \times \; -->2m{u}^0=6m{u}^0$
Is any of this correct, or is it all wrong?

I have the following task from my textbook:
Let ${\left\{f_n\right\}}_{n\in <!--\; \in \; -->\mathbb{N}}$ be a sequence of measurable function on M with
$f_n\to <!--\; \to \; -->f\text{ a.s., }$
where f is also a measurable function. Here, a.s. means almost surely (= almost everywhere).
Show: if there exists a nonnegative measurable function g satisfying the following conditions:
$\left|f_n\right|\le <!--\; \le \; -->g\text{ a.s. for all }n\in <!--\; \in \; -->\mathbb{N}$
and
${\int <!--\; \int \; -->}_M{g}^{p}d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}\text{ for one }p>0,$
then
${\int <!--\; \int \; -->}_M{|f_n-<!--\; -\; -->f|}^{p}d\mathrm{\mu <!--\; \mu \; -->}\to <!--\; \to \; -->0\text{ for }n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$
These notations remind me a little bit of the Hölder inequality.
What I have got:
I thought that we have
$|f_n{|}^p\le <!--\; \le \; -->g^{p}a.s.$
Thus
${\int <!--\; \int \; -->}_M|f_n{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_M{g}^{p}d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->infty$
And as
$f_n\to <!--\; \to \; -->f$
then the rest follows?

I wish to determine all functions $f$ of bounded variation on [0,1] such that
$f(x)+(TV(f_{[0,x]}){)}^{1/2}=1\text{for all }x\in <!--\; \in \; -->[0,1]\text{and}{\int <!--\; \int \; -->}_0^{1}f(x)dx=1/3.$
Here $TV(f_{[a,b]})$ denotes the total variation of $f$ on [a,b].
So looking at what needs to be done, it is clear to me that the function f(x) = 1/3 satisfies the integral equation however, I have no intuition on how to proceed with the equation that involves the total variation or how to bring it all together.

Let $O=\{(x_1,x_2)\in <!--\; \in \; -->{\mathbb{R}}^2:0<x_1=x_2<1\}$, $I=\{N\in <!--\; \in \; -->\mathcal{B}:\mathrm{\lambda <!--\; \lambda \; -->}((0,1)\cap <!--\; \cap \; -->N)=0\}$, where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the Lebesgue measure and $\mathcal{B}$ is the Borel sigma algebra.
Furthermore let $\mathrm{\mu <!--\; \mu \; -->}(A)=\{\begin{array}{ll}0& \mathrm{\exists <!--\; \exists \; -->}{A}_1,A_2\in <!--\; \in \; -->I,\text{ and }\mathrm{\exists <!--\; \exists \; -->}{B}_1,B_2\in <!--\; \in \; -->\mathcal{B},\text{ so }A\subseteq <!--\; \subseteq \; -->(A_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(A_2\times <!--\; \times \; -->B_2)\\ 1& \text{otherwise}\end{array}$
Show that $\mathrm{\mu <!--\; \mu \; -->}(O)=1$.
My thought is a contradiction argument. If $\mathrm{\mu <!--\; \mu \; -->}(O)=0$, then there must exist $N_1,N_2\in <!--\; \in \; -->I$ and B1,B2∈B such that $O\subseteq <!--\; \subseteq \; -->(N_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(N_2\times <!--\; \times \; -->B_2)$. But since $O$ is a line from the point (0,0) to the point (1,1), there cannot be a set with Lebesguemeasure 0 on the interval (0,1) on either axis. There is therefore no way to "describe" the line, because it goes from 0 to 1 on both axis, and therefore there does not exist $N_1,N_2\in <!--\; \in \; -->I$ so $O\subseteq <!--\; \subseteq \; -->(N_1\times <!--\; \times \; -->B_1)\cup <!--\; \cup \; -->(N_2\times <!--\; \times \; -->B_2)$
My problem is that I understand why is the measure $\mathrm{\mu <!--\; \mu \; -->}(O)=1$, I just can't seem to find the right proof technique.

If we have a question where we have to find the coefficient's units such as K in this case. The actual formula contains more parts but it is simply the derivatives that I am unsure about.
$K=\frac{dv}{dt}$
Where $v=m{s}^{-<!--\; -\; -->1}$ (metres per second) and t is time
Do I simply find the derivative of $m{s}^{-<!--\; -\; -->1}$ in respect to s? And whatever units of measurement I am left with are the units of K? ie.
$\frac{d(m{s}^{-<!--\; -\; -->1})}{ds}$
$=m\ast <!--\; \ast \; -->\frac{d(s^{-<!--\; -\; -->1})}{ds}$
$=-<!--\; -\; -->m{s}^{-<!--\; -\; -->2}$

Let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},\mathrm{\mu <!--\; \mu \; -->})$ be a measure space, with $\mathrm{\mu <!--\; \mu \; -->}$ a finite measure. Let $X:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->\mathbb{R}$ be a measurable function. Suppose
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{|X|>n}|X|\text{d}\mathbb{P}=0$
In proving the integrability of $X$, the following step is made. Take $N\in <!--\; \in \; -->\mathbb{N}$, such that ${\int <!--\; \int \; -->}_{|X|>N}|X|\text{d}\mathbb{P}\le <!--\; \le \; -->1$.
Now:
${\int <!--\; \int \; -->}_{\mathbb{R}}|X|\text{d}\mathbb{P}={\int <!--\; \int \; -->}_{|X|>N}|X|\text{d}\mathbb{P}+{\int <!--\; \int \; -->}_{|X|\le <!--\; \le \; -->N}|X|\text{d}\mathbb{P}\le <!--\; \le \; -->1+N\cdot <!--\; \cdot \; -->\mathrm{\mu <!--\; \mu \; -->}(|X|\le <!--\; \le \; -->N)$
Where $\mathrm{\mu <!--\; \mu \; -->}(|X|\le <!--\; \le \; -->N):=\mathrm{\mu <!--\; \mu \; -->}(\{\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}:|X(\mathrm{\omega <!--\; \omega \; -->})|\le <!--\; \le \; -->N\})$. Why is it that the second integral can be bounded by $N\cdot <!--\; \cdot \; -->\mathrm{\mu <!--\; \mu \; -->}(|X|\le <!--\; \le \; -->N)$?
My first thought was by somehow using the Markov inequality, but this doesn't seem to work.

I am trying to find $n$-dimensional Lebesgue measure of $A$. Let $A\subset <!--\; \subset \; -->{\mathbb{R}}^n$ be a set:
$A=\{(x_1,x_2,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{R}}^n:\left(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|x_i|\right)\le <!--\; \le \; -->1\}$
1. Exact same problem. The same problem appears here:
"Let n $\ge <!--\; \ge \; -->$ 1 and n is an integer. Use Fubini's theorem to calculate the n-th Lebesgue measure of this set:
$\{(x_1,x_2,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{R}}^n:x_i\ge <!--\; \ge \; -->0,i=1,2,\dots <!--\; \dots \; -->,\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i\le <!--\; \le \; -->1\}$"
1.1 For $n=1$, we have $0\le <!--\; \le \; -->\mathrm{\lambda <!--\; \lambda \; -->}(A)\le <!--\; \le \; -->1⟹<!--\; ⟹\; -->\mathrm{\lambda <!--\; \lambda \; -->}(A)\le <!--\; \le \; -->1$. It seems to be obvious to me.
1.2 For $n=2$, we have $0\le <!--\; \le \; -->\mathrm{\lambda <!--\; \lambda \; -->}(A)\le <!--\; \le \; -->2⟹<!--\; ⟹\; -->\mathrm{\lambda <!--\; \lambda \; -->}(A)\le <!--\; \le \; -->2$. Sqaure with $a=\sqrt{2}$
2. Similar problem. I have noticed similar problem, where A:
$A=\{(x_1,x_2,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{R}}^n:\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|x_i|:\mathrm{\forall <!--\; \forall \; -->}|x_i|\le <!--\; \le \; -->1\}.$

Reading about random sets, I've come across the phrase "$\mathcal{B}(\mathcal{F})$ is the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra generated by the topology of closed convergence." where $\mathcal{F}$ is the family of closed sets in $\mathbb{R}.$
I don't know a lot about topology, so I'm not sure how to understand/interpret this. In general, a Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on a set $S$ is the $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra generated by the open sets of $S$, and I suppose a topology gives a notion of "openness", but I would appreciate if someone could provide me some intuition behind this particular σ-algebra and the "topology of closed convergence".

I recently got this peculiar interview question, and I wanted some help figuring out how to reach an appropriate solution. Imagine that we have a race car that is driving on a 50-mile-long race track, and this race car has five minutes to drive along this race track. Suppose that I went 20 miles per hour on the first half of the race track. How fast do I need to go on the second half of the race track such that I average 40 miles per hour over the whole drive on the track?
I immediately went for the idea that the answer was 60 miles per hour, but supposedly that was wrong. I think I needed to better consider the fact that miles per hour is a measure of distance over time. So
$40\text{mph}=\frac{40\text{ miles}}{60\text{ minutes}},$
But I am now stuck on how to use this information to deduce how many minutes I need to take on the second half to average this speed. Any suggestions?

I hope this is the right place to ask this question...
My question regards the motivations behind forming numbers and their connection to physical objects. What I am not asking about, is a rigorous formulation of the natural numbers. Instead, I am curious about the perhaps more philosophical step before. Let me explain:
Generally, when we want to construct the integers (and eventually reals) we motivate the task by saying that a number represents a collection of items, addition of these numbers represents the combination of these items and so forth. (I guess this is sounding a lot like set theory as well). My question is how are these fundamental connections between numbers and the physical task of counting rigorously constructed. The way I have written it (the stuff in italics) is riddled with mysterious language such as “collection,” “combination” etc.
Of course in a purely mathematical context, this is unimportant. However, I think it’s still valuable to appreciate how abstract concepts in numbers all still hail from the mundane (yet seemingly hard to define) task of counting things.
*often times when we think of the physical models behind mathematical results, we arrive at this simple notion of “counting objects. Is this really the most precise way we can define this connection" *
For example, we “count” a collection of five apples as, well, five apples...

Let $X$ be a set and $\mathcal{F}$ a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra. Does there exist a topological space $U$ and a map $f:X\to <!--\; \to \; -->U$ such that $f$ is $\mathcal{F},\mathcal{B}$-measurable and $\mathrm{\sigma <!--\; \sigma \; -->}(f)=\mathcal{F}$? Here $\mathcal{B}$ is the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra of $U$.
Of course, this is trivial if every $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on a set is the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra with respect to some topology on the set. But this needn't be true. This is a weaker problem.

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document We have the outer measure

The inner measure is:

Where ${\mathrm{\mu <!--\; \mu \; -->}}_0(A)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_0(A_n)$ and ${\mathcal{A}}_0$ is a $\mathrm{\sigma <!--\; \sigma \; -->}$ algebra.
But why is ${\mathrm{\mu <!--\; \mu \; -->}}_{\ast <!--\; \ast \; -->}\le <!--\; \le \; -->{\mathrm{\mu <!--\; \mu \; -->}}^{\ast <!--\; \ast \; -->}$?
I always end up with a contradiction.
My proof would be:

Because:

We get:


Because by definition ${\mathrm{\mu <!--\; \mu \; -->}}^{\ast <!--\; \ast \; -->}(A)=inf\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_0(A_n)\le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_0(A_n)$ this means, that ${\mathrm{\mu <!--\; \mu \; -->}}_0(X\mathrm{\setminus <!--\; \setminus \; -->}A)-<!--\; -\; -->{\mathrm{\mu <!--\; \mu \; -->}}^{\ast <!--\; \ast \; -->}(X\mathrm{\setminus <!--\; \setminus \; -->}A)\ge <!--\; \ge \; -->0$
We get:


So where's my mistake?

I have to see if
$F(t)={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\sin <!--\; \; -->(tx)e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx$
is well defined when $t>0$ and when $t<0$.
For it, I have to see if $|F(t)|<\mathrm{\infty <!--\; \infty \; -->}$. I have done the following
$\begin{array}{rl}|<!--\; |\; -->F(t)|<!--\; |\; -->& =|{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\sin <!--\; \; -->(tx)e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx|\\ & \le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}|\frac{\sin <!--\; \; -->(tx)e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}|dx\\ & ={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{|<!--\; |\; -->\sin <!--\; \; -->(tx)|<!--\; |\; -->e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx\\ & ={\int <!--\; \int \; -->}_0^1\frac{|<!--\; |\; -->\sin <!--\; \; -->(tx)|<!--\; |\; -->e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx+{\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{|<!--\; |\; -->\sin <!--\; \; -->(tx)|<!--\; |\; -->e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx\\ & \le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^1\frac{t{e}^{-<!--\; -\; -->x^{2}t}}{x^{1/2}}dx+{\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{e^{-<!--\; -\; -->x^{2}t}}{x^{3/2}}dx\end{array}$

1. When $t>0$: The first integral
${\int <!--\; \int \; -->}_0^1\frac{t{e}^{-<!--\; -\; -->x^{2}t}}{x^{1/2}}dx\le <!--\; \le \; -->t{\int <!--\; \int \; -->}_0^1\frac{1}{x^{1/2}}dx=2t<\mathrm{\infty <!--\; \infty \; -->}$
(I can say that it is finite, right?). But the second one?2. And when $t<0$, what can we say?

I don't know how to bound them.

Let $(X,M,\mathrm{\mu <!--\; \mu \; -->})$ be a measure space s.t there exists an infinite sequence of disjoints measurable sets of strictly positive finite measure. Show that $L^1(X,M,\mathrm{\mu <!--\; \mu \; -->})$ is not reflexive.
My attempt: this question followed a question that asked to prove that a TVS $V$ is reflexive $⟺<!--\; ⟺\; -->$ the closed unit ball in $V$ is weakly compact. I tried showing that $\{f\in <!--\; \in \; -->L^1:\Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}_1\le <!--\; \le \; -->1\}$ is not weakly compact, but I wasn't able to find an open covering by weak open sets s.t it has no finite subcovering. Any help would be appreciated.

This seems trivial but I just want to be careful about it. By definition, I would like $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}P(|X_n-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}|>\mathrm{ϵ<!--\; ϵ\; -->})=0$ for $\mathrm{ϵ<!--\; ϵ\; -->}>0$. But can I just rewrite it as $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}P(|X_n-<!--\; -\; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\mathrm{\mu <!--\; \mu \; -->}}_n|>\mathrm{ϵ<!--\; ϵ\; -->})=0$ and bring the limit out from there? That seems really odd to me.

My class notes define a signed measure on a measurable space $(X,\mathcal{R})$ as a $\mathrm{\sigma <!--\; \sigma \; -->}$-additive function $\mathrm{\nu <!--\; \nu \; -->}:\mathcal{R}\to <!--\; \to \; -->\mathbb{R}$. (I take this to mean we're only considering finite measures.) I'm confused on what I'm supposed to interpret $\mathrm{\sigma <!--\; \sigma \; -->}$-additive to mean here. My first guess would be that if $A=\underset{n}{\bigcup <!--\; \bigcup \; -->}{A}_n$ where $A_1,A_2,\dots <!--\; \dots \; -->$ are disjoint, then
$\mathrm{\nu <!--\; \nu \; -->}(A)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n);$
i.e., just the usual definition of $\mathrm{\sigma <!--\; \sigma \; -->}$-additivity. But this seems problematic when the series on the right doesn't converge absolutely, because then its value depends on the order of the $A_i$, which my gut tells me shouldn't be the case. Does $\mathrm{\sigma <!--\; \sigma \; -->}$-additivity here also involve the claim that the series on the right always converges absolutely? The wikipedia page on signed measures does require this, but no other source I found online, or my class notes, explicitly states it.

Given $f_k\in <!--\; \in \; -->L^p(\mathrm{\Omega <!--\; \Omega \; -->})$, I have to prove that ${\Vert \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{f}_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}⩽<!--\; ⩽\; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}$, where $\Vert f\Vert \doteq <!--\; \doteq \; -->{\left({\displaystyle {\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}{\left|f\right|}^p}\right)}^{1/p}$ denotes the p-norm, with $1⩽<!--\; ⩽\; -->p<\mathrm{\infty <!--\; \infty \; -->}$. Since I've already proved the original Minkowski inequality, a simple induction estabilishes the finite case, i.e., ${\Vert \underset{n=1}{\overset{m}{\sum <!--\; \sum \; -->}}{f}_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}⩽<!--\; ⩽\; -->\underset{n=1}{\overset{m}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}(\ast <!--\; \ast \; -->)$, $\mathbb{N}\ni <!--\; \ni \; -->m<\mathrm{\infty <!--\; \infty \; -->}$.
So, what I've thought is, since $L^p(\mathrm{\Omega <!--\; \Omega \; -->})$ is a Banach space, to suppose that $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\Vert f_n\Vert }_{L^p(\mathrm{\Omega <!--\; \Omega \; -->})}<\mathrm{\infty <!--\; \infty \; -->}$, which would give us that $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{f}_n$ converges in $L^p(\mathrm{\Omega <!--\; \Omega \; -->})$, by completeness. Hence, the desired inequality follow by making $m⟶<!--\; ⟶\; -->\mathrm{\infty <!--\; \infty \; -->}$ in $(\ast <!--\; \ast \; -->)$, but I'm not completely sure about this.
Any help will be appreciated.

Let $(\mathsf{X},\mathcal{X},\mathrm{\mu <!--\; \mu \; -->})$ be a measure space with $\mathrm{\mu <!--\; \mu \; -->}$ begin $\mathrm{\sigma <!--\; \sigma \; -->}$-finite.
Definition of Random Variable: Let $(\mathsf{Y},\mathcal{Y})$ be a measurable space. A function $\mathrm{\xi <!--\; \xi \; -->}:\mathsf{X}\to <!--\; \to \; -->\mathsf{Y}$ is a random variable if ${\mathrm{\xi <!--\; \xi \; -->}}^{-<!--\; -\; -->1}(A)\in <!--\; \in \; -->\mathcal{X}$ for every $A\in <!--\; \in \; -->\mathcal{Y}$. We say that the random variable is a $\mathcal{X}$-measurable function.
Definition of Radon-Nikodym derivative: Let $\mathrm{\lambda <!--\; \lambda \; -->}$ be $\mathrm{\sigma <!--\; \sigma \; -->}$-finite measure with $\mathrm{\mu <!--\; \mu \; -->}\ll <!--\; \ll \; -->\mathrm{\lambda <!--\; \lambda \; -->}$. Then there exists a $\mathcal{X}$-measurable function $f:\mathsf{X}\to <!--\; \to \; -->[0,+\mathrm{\infty <!--\; \infty \; -->})$ denoted
$f=\frac{d\mathrm{\mu <!--\; \mu \; -->}}{d\mathrm{\lambda <!--\; \lambda \; -->}}$
satisfying
$\mathrm{\mu <!--\; \mu \; -->}(A)={\int <!--\; \int \; -->}_{A}fd\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\forall <!--\; \forall \; -->}A\in <!--\; \in \; -->\mathcal{X}.$
It seems that the Radon-Nikodym derivative is a $\mathcal{X}$-measurable function so by definition it should be a random variable between $(\mathsf{X},\mathcal{X})$ and $({\mathbb{R}}_{\ge <!--\; \ge \; -->0},\mathcal{B}({\mathbb{R}}_{\ge <!--\; \ge \; -->0}))$. Is this the case that every Radon-Nikodym derivative is a random variable?

Let us say that we have some set of independent random variables: $X=\{X_i{\}}_{i=1}^n$ defined over a probability space of $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},P)$. I want to understand whether the following holds:
If ${\mathcal{F}}_X=\mathrm{\sigma <!--\; \sigma \; -->}\left(\{X_1,X_2,\dots <!--\; \dots \; -->,X_n\}\right)$, then is it, in general, true that: $\left|{\mathcal{F}}_X\right|=\underset{\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->I_n}{\sum <!--\; \sum \; -->}|\mathrm{\sigma <!--\; \sigma \; -->}(\{X_i\})|$?
As we know, as $X$ consists of independent R.V.-s, then we can state that $\mathrm{\forall <!--\; \forall \; -->}i,j\in <!--\; \in \; -->I_n:\mathrm{\sigma <!--\; \sigma \; -->}(\{X_i\})$ and $\mathrm{\sigma <!--\; \sigma \; -->}(\{X_j\})$ are independent. But how to proceed from this fact to the split of the cardinality of ${\mathcal{F}}_X$?
I would appreciate any help, thank you in advance!