Explore Measurement Concepts with Our Examples and Questions

1 kg-2.2 pounds (ibs for #)
1 ounce-28.35 (30) g
Again, you will use these constantly. Memorize these conversions
How many grams in pounds?
Practice:
A patient chart lists her weight as 150#. How many kg?
A man is listed as 100 kg. How many pounds?

Draw, in standard position, the angle whose measure is given:
${210}^{\circ <!--\; \circ \; -->}$

I'm given a random variable $X_n$ that is wrt the following measure:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\mathrm{\#<!--\; \#\; -->}\{X_n\in <!--\; \in \; -->[a,b]\}}{n}={\int <!--\; \int \; -->}_a^b{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}.$
with an error term $O(\log <!--\; \; -->n/n)$.
Now I'm trying to find the distribution of ${\cos }^2<!--\; \; -->X_n$. The text I'm reading says this is ${\int <!--\; \int \; -->}_a^b{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}$, which I'm already not sure of, and I also want to figure out what the error term of this is.

Let B([0,1]) be the borel set on [0,1] and suppose that $\mathrm{\mu <!--\; \mu \; -->}$ is a finite measure on B([0,1]). Then, define $A=[0,1]\cap <!--\; \cap \; -->\{x;\mathrm{\mu <!--\; \mu \; -->}(\left\{x\right\})>0\}$. I have shown that if A is countable, then [0,1]∖A is a separable metric space (because it is a subset of a separable metric space). Now if $(x_n{)}_n$ is a dense sequence in [0,1]∖A, then I want to show that it is also dense in [0,1]. How can I prove it?

This is from Durrett's probability:
Let $X_n\in <!--\; \in \; -->[0,1]$ be adpated to ${\mathcal{F}}_n$. Let $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0$ with $\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->}=1$ and suppose
$\mathbb{P}(X_{n+1}=\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->}{X}_n\mid <!--\; \mid \; -->{\mathcal{F}}_n)=X_n,\mathbb{P}(X_{n+1}=\mathrm{\beta <!--\; \beta \; -->}{X}_n\mid <!--\; \mid \; -->{\mathcal{F}}_n)=1-<!--\; -\; -->X_n$
Show $\mathbb{P}(\underset{n}{lim}{X}_n=0\text{or}1)=1.$
I know there is question Martingale convergence proof here but can't we prove this using levy's 0-1 law?
What can we say about levy's 0-1 law regarding Martingale Sequence?

I have some problems, help:
Let $(X,{\mathcal{A}}_{\mathcal{1}},\mathrm{\nu <!--\; \nu \; -->})$ be a probability space. Let $A:L_1\to <!--\; \to \; -->L_1$ be a contraction such that $A:L_{\mathrm{\infty <!--\; \infty \; -->}}\to <!--\; \to \; -->L_{\mathrm{\infty <!--\; \infty \; -->}}$ is also a contraction. Suppose $A$ preserves positivity.
In a proof of why $A:L_p\to <!--\; \to \; -->L_p$ is a contraction for every $1\le <!--\; \le \; -->p<\mathrm{\infty <!--\; \infty \; -->}$, my lecture notes state that for all $f\ge <!--\; \ge \; -->0$, we have
$\Vert <!--\; \Vert \; -->Af{\Vert <!--\; \Vert \; -->}_p=\underset{g\ge <!--\; \ge \; -->0,\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}_q=1}{sup}{\int <!--\; \int \; -->}_X(Af\cdot <!--\; \cdot \; -->g),$
where $\frac{1}{p}+\frac{1}{q}=1$.
Why does the above equality hold? The strangest part to me is the fact that the LHS has a power of $\frac{1}{p}$ outside of the integral ${\int <!--\; \int \; -->}_X|Af{|}^p$, but in the RHS the integral is not being raised to any power. I've tried to raise both sides to $p$ but I could not do much with this. If it helps, this came up in the study of conditional expectations.

I measured something $N$ times using different measurement techniques. Each measurement technique $i$ has a known variance ${\mathrm{\sigma <!--\; \sigma \; -->}}_i^2$. So every measurement is $x_i=\stackrel{^<!--\; ^\; -->}{x}+{\mathrm{ϵ<!--\; ϵ\; -->}}_i$ where $\stackrel{^<!--\; ^\; -->}{x}$ is the true value and ${\mathrm{ϵ<!--\; ϵ\; -->}}_i$ is pulled from a normal distribution with mean 0 and variance ${\mathrm{\sigma <!--\; \sigma \; -->}}_i^2$.
Ok, I have all these measurements. Now I want to know the variance of my measurements taken together, taking into account the fact that I know the variance of all the measurement techniques. What's the best way to do this?

I am trying to find the %Accuracy, in which I used this equation: %Accuracy = 100-%Error.
So far, I have faced two problems with this equation:
1. When the Exact Value is Zero, the fraction can’t be used -> Solved by adding the same value to both Exact and Measured, to avoid the null denominator
2. When the Measured value is twice bigger or smaller, the %Accuracy value will be in negative values, In which I don't see the meaning behind it.

For example:
if: Exact = 20; Measured = 25; %Accuracy = 75%;
But, when: Exact = 20; Measured = 45; %Accuracy = -25%;
What is the meaning of -25%? and How to change the range of [0-100], to take the values that are outside of it accurately?

Absolute continuity: For all $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ there exists a $\mathrm{\delta <!--\; \delta \; -->}>0$ such that if $\mathrm{\mu <!--\; \mu \; -->}(A)<\mathrm{\delta <!--\; \delta \; -->}$, ${\int <!--\; \int \; -->}_A|f|d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{ϵ<!--\; ϵ\; -->}$. Here $A$ is a measurable subset of $E$.
I know that if $f$ is integrable then it is absolutely continuous. But is there anyway I can show that absolute continuity implies integrability when $\mathrm{\mu <!--\; \mu \; -->}$ is finite?
$\int <!--\; \int \; -->|f|d\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_A|f|d\mathrm{\mu <!--\; \mu \; -->}+{\int <!--\; \int \; -->}_{A^c}|f|d\mathrm{\mu <!--\; \mu \; -->}$
Can I choose some special set $A$ such that $\mathrm{\mu <!--\; \mu \; -->}(A)<\mathrm{\delta <!--\; \delta \; -->}$?

"Let ${\mathrm{\mu <!--\; \mu \; -->}}_0(A)=\mathrm{\mu <!--\; \mu \; -->}(A\cup <!--\; \cup \; -->A_0),A\in <!--\; \in \; -->\mathcal{F}$ be a measure. Show that if
$\int <!--\; \int \; -->fd\mathrm{\mu <!--\; \mu \; -->}$
exists, then
${\int <!--\; \int \; -->}_{A_0}fd\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->f{d}_{{\mathrm{\mu <!--\; \mu \; -->}}_0}$"
However, I think there's a typo in this exercise. ${\mathrm{\mu <!--\; \mu \; -->}}_0$ isn't even a measure when $\mathrm{\mu <!--\; \mu \; -->}(A_0)\ne <!--\; \ne \; -->0$, since ${\mathrm{\mu <!--\; \mu \; -->}}_0(\mathrm{\varnothing <!--\; \varnothing \; -->})=\mathrm{\mu <!--\; \mu \; -->}(A_0)$. If it were ${\mathrm{\mu <!--\; \mu \; -->}}_0(A)=\mathrm{\mu <!--\; \mu \; -->}(A\cap <!--\; \cap \; -->A_0)$, which I think that's what the author meant, I could prove it in the following way:
For simple functions,
${\int <!--\; \int \; -->}_{A_0}fd\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->f{\mathbf{1}}_{A_0}d\mathrm{\mu <!--\; \mu \; -->}=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{f}_i{\mathbf{1}}_{A_0}\mathrm{\mu <!--\; \mu \; -->}(A_i)=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{f}_i\mathrm{\mu <!--\; \mu \; -->}(A_i\cap <!--\; \cap \; -->A_0)=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{f}_i{\mathrm{\mu <!--\; \mu \; -->}}_0(A_i)=\int <!--\; \int \; -->fd{\mathrm{\mu <!--\; \mu \; -->}}_0$
For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $(s_n{)}_{n\in <!--\; \in \; -->\mathbb{N}}$ such that $s_n\to <!--\; \to \; -->f$, then
${\int <!--\; \int \; -->}_{A_0}fd\mathrm{\mu <!--\; \mu \; -->}=\int <!--\; \int \; -->f{\mathbf{1}}_{A_0}d\mathrm{\mu <!--\; \mu \; -->}=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\int <!--\; \int \; -->s_n{\mathbf{1}}_{A_0}d\mathrm{\mu <!--\; \mu \; -->}=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\int <!--\; \int \; -->s_{n}d{\mathrm{\mu <!--\; \mu \; -->}}_0=\int <!--\; \int \; -->fd{\mathrm{\mu <!--\; \mu \; -->}}_0$
And finally, for any measurable function, we just use $f=f^{+}-<!--\; -\; -->f^{-<!--\; -\; -->}$. So, which one is it, $\mathrm{\mu <!--\; \mu \; -->}(A\cup <!--\; \cup \; -->A_0)$ or $\mathrm{\mu <!--\; \mu \; -->}(A\cap <!--\; \cap \; -->A_0)$?

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Suppose $E$ is a measurable set and $m(E)<+\mathrm{\infty <!--\; \infty \; -->}$. $\{f_k\}$ is a sequence of non-negative measurable functions. If

prove that $f_k$ converge to 0 in measure.

If we can prove $\frac{f_k(x)}{1+f_k(x)}\to <!--\; \to \; -->0$ a.e.$x\in <!--\; \in \; -->E$, then the problem is easy to solve. Is it correct? If so, how to prove that?
Appreciate any hint or help!

Let $X$ be a compact Hausdorff space and $\mathrm{\mu <!--\; \mu \; -->}$ be a complex Borel measure on $X$ such that ${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}\ge <!--\; \ge \; -->0$ for every $f\in <!--\; \in \; -->C(X)$ with $f\ge <!--\; \ge \; -->0.$ Then show that $\mathrm{\mu <!--\; \mu \; -->}$ is non-negative.
Could anyone give some suggestion in this regard?

This could totally be a stupid question but I'm unsure: is a measure (ie positive, countable additive on a $\mathrm{\sigma <!--\; \sigma \; -->}$ algebra, 0 for the empty set) actually a measurable function (wrt to the Borel-sigma algebra on $\mathbb{R}$)?

We have known that in general the smallest field containing $k$ subsets has, in the general case, size $2^{2^k}$.
However, in this question, follows the above result, we will have $2^{2^3}=256$ elements in the smallest field $\mathcal{F}$. How can we represent it?

Let $M\subset <!--\; \subset \; -->\mathbb{R}$. Is $M\times <!--\; \times \; -->M\times <!--\; \times \; -->\mathbb{Q}\subset <!--\; \subset \; -->{\mathbb{R}}^3$ measurable? (Lebesgue measurable)
I would say no, because by the definition the the Lebesgue measure, intuitively speaking, what it does is:
We "approximate" our function by continuous functions with a compact support. Then we take the Riemann integral of that in each dimension.
So if M is non-measurable it seems to me that then $M\times <!--\; \times \; -->M\times <!--\; \times \; -->\mathbb{Q}$ is too (so e.g. let M be the Vitali set).
On the other hand, I think you could apply Tonelli Fubini:
${\int <!--\; \int \; -->}_M^{\ast <!--\; \ast \; -->}{\int <!--\; \int \; -->}_M^{\ast <!--\; \ast \; -->}{\int <!--\; \int \; -->}_{\mathbb{Q}}^{\ast <!--\; \ast \; -->}{1}_{M\times <!--\; \times \; -->M\times <!--\; \times \; -->\mathbb{Q}}={\int <!--\; \int \; -->}_M^{\ast <!--\; \ast \; -->}{\int <!--\; \int \; -->}_M^{\ast <!--\; \ast \; -->}0$
since the measure of $\mathbb{Q}$ is 0. So the above integral is also 0, so we can apply Tonelli, thus Fubini, and thus $1_{M\times <!--\; \times \; -->M\times <!--\; \times \; -->\mathbb{Q}}$ is integrable and hence measurable.
Is this correct? So my intuitive understanding is wrong, and instead it should be something like "since we can choose the order of the Riemann integrals, we can first do the one that results in zero, and then it doesn't matter that the other 'dimensions' aren't measurable anymore, since we have 0 everywhere"?

I'm studying Pavliotis' Stochastic Processes and I am having trouble with one of the exercises.
Specifically the first exercise of chapter two is prove that the Markov property in the sense off the immediate future being independent of the past given the present, i.e. $P(X_{n+1}|X_1,\dots <!--\; \dots \; -->,X_n)=P(X_{n+1}|X_n)$ implies that arbitrary futures are independent of the past given the present, i.e. $P(X_{n+m}|X_1,\dots <!--\; \dots \; -->,X_n)=P(X_{n+m}|X_n)$.
Conceptually I imagine a proof using induction along the lines of: Assume the equality holds for some m, then
$\begin{array}{c}P(X_{n+m+1}|X_1,\dots <!--\; \dots \; -->,X_n)=\int <!--\; \int \; -->P(X_{n+m+1},X_{n+m}=x|X_1,\dots <!--\; \dots \; -->,X_n)dx\\ =\int <!--\; \int \; -->P(X_{n+m+1}|X_{n+m}=x)P(X_{n+m}=x|X_1,\dots <!--\; \dots \; -->,X_n)dx\\ =\int <!--\; \int \; -->P(X_{n+m+1}|X_{n+m}=x)P(X_{n+m}=x|X_n)dx\\ =\int <!--\; \int \; -->P(X_{n+m+1},X_{n+m}=x|X_n)dx=P(X_{n+m+1}|X_n)\end{array}$
However, this is based on intuition and my experience non-measure-theoretic probability theory. I cannot justify these steps when I think of $P$ as a measure and $P(X=x)$ meaning $P\{X^{-<!--\; -\; -->1}(x)\}$. In fact I have had trouble finding a definition of conditional probability that is at all helpful. Most authors seem to provide the abstract definition in terms of conditional expectation with respect to $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra, but I haven't found any resources that show how to work with this definition.
So my question is: how (if it all) are these steps, particularly the first two equalities justified from a measure theoretic perspective?

My proof is as follow (S is the event space in the probability space of an experiment):
If $A\subset <!--\; \subset \; -->B\in <!--\; \in \; -->S$ then $A^c\cap <!--\; \cap \; -->B\in <!--\; \in \; -->S$ so this means that $A^c\cap <!--\; \cap \; -->B$ contains B but not A. This implies that $B\setminus <!--\; \setminus \; -->A\in <!--\; \in \; -->S$
I'm not sure if this prove is correct and if the rigor is strong enough.

Let $f:{\mathbb{R}}^k\to <!--\; \to \; -->\mathbb{R}$ be a bounded harmonic function (i.e. $f$ is of class $C^2$ and $\underset{r=1}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}f}{\mathrm{\partial <!--\; \partial \; -->}{y}_r^2}=0$) and $(x_1,...,x_k)\in <!--\; \in \; -->{\mathbb{R}}^k.$
Prove that
$f(x_1,...,x_k)=\frac{1}{(\sqrt{2\mathrm{\pi <!--\; \pi \; -->}}{)}^k}{\int <!--\; \int \; -->}_{{\mathbb{R}}^k}f(x_1+y_1,...,x_k+y_k)e^{-<!--\; -\; -->\frac{1}{2}\underset{r=1}{\overset{k}{\sum <!--\; \sum \; -->}}{y}_r^2}d{y}_1...d{y}_k.$
We note that
${\int <!--\; \int \; -->}_{{\mathbb{R}}^k}f(x_1+y_1,...,x_k+y_k)e^{-<!--\; -\; -->\frac{1}{2}\underset{r=1}{\overset{k}{\sum <!--\; \sum \; -->}}{y}_r^2}d{y}_1...d{y}_k={\int <!--\; \int \; -->}_{{\mathbb{R}}^k}f(y_1,...,y_k)e^{-<!--\; -\; -->\frac{1}{2}\underset{r=1}{\overset{k}{\sum <!--\; \sum \; -->}}(y_r-<!--\; -\; -->x_r{)}^2}d{y}_1..d{y}_k,$
how to use the fact that $f$ is harmonic ?

Consider the function $f(x):[0,1]\to <!--\; \to \; -->[0,1]$ given by
$\{\begin{array}{ll}2x& 0\le <!--\; \le \; -->x\le <!--\; \le \; -->\frac{1}{2}\\ x-<!--\; -\; -->\frac{1}{2}& \frac{1}{2}<x\le <!--\; \le \; -->1\end{array}$
I found the measure with density given by $\mathrm{\rho <!--\; \rho \; -->}=\frac{4}{3}{\mathrm{\chi <!--\; \chi \; -->}}_{[0,\frac{1}{2}]}+\frac{2}{3}{\mathrm{\chi <!--\; \chi \; -->}}_{(\frac{1}{2},1]}$ (with respect to Lebsegue measure) is invariant for this transformation. My question now is: how can I prove this system with this measure is ergodic? I thought to use the approach with invariant functions and Fourier series, but I'm not sure on how to write Fourier expansion with a measure different than Lebesgue's. I also thought to exploit a possible conjugacy with symbolic shift, but wasn't able to prove that $[0,\frac{1}{2}]$ and $(\frac{1}{2},1]$ constitute a Markov partition of the unit interval. Any ideas?

Let's say we're given a Lebesgue-Stieltjes measure $\mathrm{\mu <!--\; \mu \; -->}$ and an associated function $F_{\mathrm{\mu <!--\; \mu \; -->}}$ such that
$\mathrm{\mu <!--\; \mu \; -->}([a,b))=F_{\mathrm{\mu <!--\; \mu \; -->}}(b)-<!--\; -\; -->F_{\mathrm{\mu <!--\; \mu \; -->}}(a)$
for all semi-intervals in $\mathbb{R}.$
I aim to calculate $\mathrm{\mu <!--\; \mu \; -->}([a,b]),\mathrm{\mu <!--\; \mu \; -->}((a,b)),\mathrm{\mu <!--\; \mu \; -->}((a,b])$ in terms of the function $F_{\mathrm{\mu <!--\; \mu \; -->}}.$
My current idea is to express the closed, open, and semi-interval in terms of the left semi-interval as such:
$(a,b)=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}[a_i,b)$
where $a<a_{i+1}<a_i$ and $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_n=a$. For example, such a sequence could be $a_i=a+2^{-<!--\; -\; -->i}.$
Then,$\mathrm{\mu <!--\; \mu \; -->}((a,b))=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\mu <!--\; \mu \; -->}([a_i,b))=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(F_{\mathrm{\mu <!--\; \mu \; -->}}(b)-<!--\; -\; -->F_{\mathrm{\mu <!--\; \mu \; -->}}(a+2^{-<!--\; -\; -->i})).$
We could get similar expressions using $(a,b]=[a-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->},b]\mathrm{╲<!--\; ╲\; -->}[a-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->},a]$ and then substituting as needed.
However, all of these expressions would be extremely messy and would give very complicated answers. Is there a cleaner way of expressing the measure of each interval?