Juan makes a measurement in a chemistry laboratory and records the result in his lab report.

When might we be likely to use a coefficient of variation?
а. when comparing measurements that are on different scales
b. when comparing measurements that are made of numeric data
c. when comparing measurements that are made of ordinal data
d. when comparing measurements that are made of grouped data

If we have $p<q$ with $p,q\in [1,\mathrm{\infty })$ and a sequence $(f_n)\in L^p\cap L^q$, $(f_n)\to f\in L^p$ in $L^p$, what type of conditions are required to determine that $f_n\to f$ in $L^q$?
My thought process so far: First and foremost, one needs that $f\in L^q$ as well. Otherwise convergence in $L^q$ is not possible. As a finite measure space implies the other direction I would assume that a finite measure space is also required. Are these two assumptions already sufficient? I am quite sure they arent, but I cannot come up with a counter example. I also tried proving that the claim is correct, but I dont see how one can approximate $|\cdot {|}_{L_q}$ with $|\cdot {|}_{L_p}$, as the first is generally bigger. Maybe it works with Holder, but I dont see how.
Any help is greatly appreciated!

Given a sigma algebra $\mathcal{S}$ over some set $X$ which is generated by some set $S\subset \mathcal{P}(X)$, and a probability function $P:\mathcal{S}\to \mathbb{R}$ why is it sufficient to check the conditions: $P(A)\ge 0,P(X)=1,P({\cup }_n{A}_n)=\sum _{n}P(A_n)$ for all $A,A_n\in S$ where $(A_n)$ is a disjoint sequence, to determine that $P$ is a valid probability? Intuitively it is obvious that if we check the conditions to be true on a generating set then we should have the conditions true for all the elements of $\mathcal{S}$, but how do we rigorously prove this fact?
For example suppose $X=\mathbb{R},\mathcal{S}=\mathcal{B}$ (the Borel sigma algebra) and $S=\{I:I\text{ is an interval in }\mathbb{R}\}$. Now it is easy to check that $P(I)=\frac{1}{\sqrt{2\pi }}{\int }_{I_x}{e}^{-x^2/2}dx$ satisfies the requisite conditions. How does it follow that the conditions are now met for $P$? Is it because of the result known as the Carathéodory's extension theorem? If so, can anyone refer me to proof, preferably in the context of probability measure?

NOMINAL, ORNIDAL, INTERVAL AND RATIO
Calender year is an example of what scale of measurement?

Need help:
Let $(\mathrm{\Omega },F,P)$ be a probability space. Suppose $f:\mathbb{R}\to \mathbb{R}$ is some function such that $f(X)$ is measurable for every real valued random variable X. I am curious if $f(X)$ is necessarily $\sigma (X)$-measurable. I tried to conclude with Doob-Dynkin lemma but to do we would need $f$ to be $\mathcal{B}(\mathbb{R})$ measurable. Does someone has an idea or is this is false in general?

I'm trying to find an answer to something.
Here $\lambda$ denotes the Lebesgue measure. Furthermore:
$f_n(x)=1_{[0,1]}(2^{m}x-j)$
Where $n=2^m+j,0\le j\le 2^m-1$. Prove that $\underset{n}{lim}{\int }^{\ast }|f_n|d\lambda \to 0$ and that $f_n(x)$ doesn't converge pointwise for any $x\in [0,1]$.
The second part isn't that difficult and not of too much interest. I am rather interested in the first part. My go-to approach would be to define a new function that does the same for $x\in [0,1]$ and from there we draw a line to zero. This would then be a function which is continuous with compact support. This means that the Lebesgue-Integral would simply be the Riemann Integral.
Let us call this function ${\varphi }_n$. The problem I run into, however, is that
${\int }^{\ast }|f_n|d\lambda ={\int }^{\ast }\underset{n}{lim}{\varphi }_{n}d\lambda ={\int }_0^1|2^{m}x-j|=|2^{m-1}-j|$
And thus we get:
${\int }^{\ast }\underset{n}{lim}|f_n|d\lambda =\underset{n}{lim}|2^{m-1}-j|\nrightarrow 0$
What is going on? I would be very grateful for your help!

Classify each of the following by type of variables and level of measurement. a. Monthly payment: $1427 b. Number of jobs in the past 10 years: 1 c. Annual family income: $86, 000 d. Marital status: Married