Explore Measurement Concepts with Our Examples and Questions

Consider a random variable $X$ with $E(X^2)<\mathrm{\infty <!--\; \infty \; -->}$. How do I show that $nP(|X|>\mathrm{\epsilon <!--\; \epsilon \; -->}\sqrt{n})\to <!--\; \to \; -->0$ for each $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$? I don't even know where to start, and would like any hint.

As $X\in <!--\; \in \; -->L^2$, $\{X\}$ is an absolutely continuous collection, wich means that there is an $N\in <!--\; \in \; -->\mathbb{N}$ such that
${\int <!--\; \int \; -->}_{|X|>N}XdP<\mathrm{\epsilon <!--\; \epsilon \; -->}$
Where can I go from there?

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let $Ч$ and $Y$ be absolutely continuous and discrete random variables, respectively. Does random vector $(X,Y)$ have to be absolutely continuous with respect to the product measure on the respective supports of $Ч$ and $Y$?
I think this is true because the following equality holds

so $f_{X,Y}(x,y)$ is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

Given that measurement along the X, Y and Z axes correspond to the terms "width", "height", and "depth", is there an accepted term for spatial measurement along the W axis when dealing in four dimensions? For that matter, are there such terms for additional dimensional axes?

Let $X$ and $Y$ be random variables. The Ky Fan metric is defined as:
$d(X,Y):=min\{\mathrm{ϵ<!--\; ϵ\; -->}>0:P(|X-<!--\; -\; -->Y|>\mathrm{ϵ<!--\; ϵ\; -->})\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}\}$
I want to show it is indeed a metric, for which I need to show that it satisfies triangle inequality.
Set $d(X,Y)={\mathrm{ϵ<!--\; ϵ\; -->}}_1,d(Y,Z)={\mathrm{ϵ<!--\; ϵ\; -->}}_2$,and $d(Z,X)={\mathrm{ϵ<!--\; ϵ\; -->}}_3$ .
I approached this by trying to show that $({\mathrm{ϵ<!--\; ϵ\; -->}}_1+{\mathrm{ϵ<!--\; ϵ\; -->}}_2)\in <!--\; \in \; -->\{\mathrm{ϵ<!--\; ϵ\; -->}>0:P(|X-<!--\; -\; -->Z|>\mathrm{ϵ<!--\; ϵ\; -->})\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}\}$. But to show that, I had to show $P(|X-<!--\; -\; -->Y|>{\mathrm{ϵ<!--\; ϵ\; -->}}_1)+P(|Y-<!--\; -\; -->Z|>{\mathrm{ϵ<!--\; ϵ\; -->}}_2)\ge <!--\; \ge \; -->P(|X-<!--\; -\; -->Z|>{\mathrm{ϵ<!--\; ϵ\; -->}}_1+{\mathrm{ϵ<!--\; ϵ\; -->}}_2)$, which I could not. Can you help in showing this?

Let $X$ be a symmetric random variable around 0 so that $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathbb{R}$ we have $\mathbb{P}(X⩽<!--\; ⩽\; -->-<!--\; -\; -->x)=\mathbb{P}(X⩾<!--\; ⩾\; -->x)$. And it's clear that the expectation value $\mathbb{E}X=0$.
My question is that, given any $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$, whether or not $\mathbb{E}[X{I}_{\{|X|<\mathrm{\epsilon <!--\; \epsilon \; -->}\}}]=0$ still holds, where $I$ is indicator function.
Intuitively I think it's right, such as taking $X\sim <!--\; \sim \; -->\mathcal{N}(0,1)$. But when I try to show it in general case, I have to proof that the following Lebesgue integral
$E[X{I}_{\{|X|<\mathrm{\epsilon <!--\; \epsilon \; -->}\}}]={\int <!--\; \int \; -->}_{|X|<\mathrm{\epsilon <!--\; \epsilon \; -->}}X\mathrm{d}\mathbb{P}=0$
and I'm stuck here. Can you help me? Thanks a lot.

Additivity of outer measure if one of the sets is open.
Suppose $A$ and $G$ are disjoint subsets of $\mathbb{R}$ and $G$ is open. Then $|A\cup <!--\; \cup \; -->G|=|A|+|G|$.
*Note that: |$A$| and l($I$) denote the outer measure of $A$ and the length of interval $I$ respectively.
I wonder that, for the part of proof to the case $G=(a,b)$ and the deduced inequality:
$\begin{array}{rl}\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}l(I_n)& =\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(l(J_n)+l(L_n))+\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}l(K_n)\\ & \ge <!--\; \ge \; -->|A|+|G|\end{array}$
How can the above inequality show that $|A\cup <!--\; \cup \; -->G|\ge <!--\; \ge \; -->|A|+|G|$?
I can only deduce from $(A\cup <!--\; \cup \; -->G)\subset <!--\; \subset \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}{I}_n$, that:
$|A\cup <!--\; \cup \; -->G|\le <!--\; \le \; -->\left|\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}{I}_n\right|\le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}l(I_n)$
However, along with:
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}l(I_n)\ge <!--\; \ge \; -->|A|+|G|$
then both the inequalities are on the same side, so I cannot link them to prove what I desire. How did author reach his conclusion?

A problem in Billingsley's Probability and Measure says that if $f:[0,1]\to <!--\; \to \; -->\mathbb{R}$ is finite-valued and Borel-measurable, then $f$ may be assumed integrable, and even bounded.

I don't see how this is true though. For example define $f$ as
$f=0\text{ on }[0,1/2)$
$f=1\text{ on }[1/2,3/4)$
$f=2\text{ on }[3/4,7/8)$
$⋮<!--\; ⋮\; -->$
$f=n\text{ on }[\frac{2n-<!--\; -\; -->1}{2^n},\frac{2n+1}{2^{n+1}})$
Then $f(x)<\mathrm{\infty <!--\; \infty \; -->}$ for all $x\in <!--\; \in \; -->[0,1]$ and $f$ is measurable since it is a step function and takes only discrete values. However $f$ is not bounded on [0,1].
What's wrong with this attempted counterexample? How would the statement be proven?

Let $(X,\mathcal{A},\mathrm{\nu <!--\; \nu \; -->})$ be a measure space. Suppose that the measure of $X$ is bounded. Show that for disjoint $A_1,A_2,\dots <!--\; \dots \; -->$ $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$.

Since $X$ is bounded from countable additivity we have that
$\mathrm{\nu <!--\; \nu \; -->}\left(\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}{A}_n\right)=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n)\le <!--\; \le \; -->\mathrm{\nu <!--\; \nu \; -->}(X)<\mathrm{\infty <!--\; \infty \; -->}.$

Thus from Borel-Cantelli
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\mathrm{\nu <!--\; \nu \; -->}(A_n)<\mathrm{\infty <!--\; \infty \; -->}⟹<!--\; ⟹\; -->\mathrm{\nu <!--\; \nu \; -->}(\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}{A}_n)=0.$

Now I also have that
$\mathrm{\nu <!--\; \nu \; -->}(\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}{A}_n)\ge <!--\; \ge \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)$
which implies that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0.$

How can I deduce from here that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$? It seems I would need to use the fact that
$0=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}\mathrm{\nu <!--\; \nu \; -->}(A_n)\ge <!--\; \ge \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim inf}\mathrm{\nu <!--\; \nu \; -->}(A_n)\ge <!--\; \ge \; -->0$
so $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim inf}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$ also and this would imply that $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\mathrm{\nu <!--\; \nu \; -->}(A_n)=0$ because of?

Let $U\subseteq <!--\; \subseteq \; -->\mathbb{R}$ be open, $f:U\to <!--\; \to \; -->\mathbb{R}$ a convex function and $m_t(x):=\frac{f(x+t)+f(x-<!--\; -\; -->t)-<!--\; -\; -->2f(x)}{2}$.
We may suppose that $U=(a,b)$ wehre $a\in <!--\; \in \; -->\mathbb{R}\cup <!--\; \cup \; -->\{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}\}$ and $b\in <!--\; \in \; -->\mathbb{R}\cup <!--\; \cup \; -->\{+\mathrm{\infty <!--\; \infty \; -->}\}$. Note that convexity of $f$ implies that $m_t(x)$ is non-negative.
Further, let $\mathbb{P}(A):={\int <!--\; \int \; -->}_A{e}^{-<!--\; -\; -->f(x)}dx$ be a probability measure.
Denote $N_t:=\{x\in <!--\; \in \; -->\mathbb{R}:m_t(x)=0\}$ and suppose that $\mathbb{P}(N_t)>\frac{1}{4}$ for all $t>0$.
Is it true (and is there a simple way to prove) that $f$ is constant on $U$?
It seems to me that this should hold but I have not been able to prove it yet.
Thank you.

I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F})$ be a measurable space, and $\mathbb{P}$ and $\mathbb{Q}$ be two probability measures on this space such that $\mathbb{Q}\ll <!--\; \ll \; -->\mathbb{P}$. Radon-Nikodym theorem tells us that there exists a $\mathcal{F}$-measurable function $f:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ denoted by $\frac{d\mathbb{Q}}{d\mathbb{P}}$, such that
$\mathbb{Q}(E)={\int <!--\; \int \; -->}_{E}f(\mathrm{\omega <!--\; \omega \; -->})d\mathbb{P}(\mathrm{\omega <!--\; \omega \; -->})$
for all $E\in <!--\; \in \; -->\mathcal{F}$.
If somehow we know the expression for $f(\mathrm{\omega <!--\; \omega \; -->})$ already, and $X:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->{\mathbb{R}}^d$ is a random vector, is there a way to derive the Radon-Nikodym derivative $\frac{d{\mathbb{Q}}^X}{d{\mathbb{P}}^X}$ for the induced distribution measures ${\mathbb{Q}}^X$ and ${\mathbb{P}}^X$ on $({\mathbb{R}}^d,\mathcal{B}({\mathbb{R}}^d))$? My rough guess was $\frac{d{\mathbb{Q}}^X}{d{\mathbb{P}}^X}(X(\mathrm{\omega <!--\; \omega \; -->}))=f(\mathrm{\omega <!--\; \omega \; -->})$ but am not sure how to prove/disprove it.

I want to implement a Kalman Filter for predicting x/y positions. I have a sensor which gives me the current position (noisy). Now I want to smooth and predict the position. Thus Kalman Filter came to my mind.

How do I have to design the filter taking the time in regards? I want to predict the next state which is ahead of the upcoming measurement. Moreover since i do not have a velocity, I'd like to estimate the velocity by the kalman as well.

State := [xpos,ypos,xvelocity,yvelocity]
Measurement := [xpos,ypos]
ControlInput := predictionTime

I would run the following algorithm, whenever i get a measurement:

1. Measure
2. Update kalman gain
3. Predict
4. Get State estimate

Now, when i use a predictionTime which is newer than the next measurement, the next measurement does not fit to the predicted state.

Is there a strategy to solve this issue? My predicted state and my next measurement do not fit, how could I fix this?

Showing that ${\int <!--\; \int \; -->}_X\log <!--\; \; -->(f)d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}(X)\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}$ given that ${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}=1$ for positive $f$ without using Jensen's inequality.
We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:
in a finite measure space $f\in <!--\; \in \; -->L_0^{+}(\mathrm{\Sigma <!--\; \Sigma \; -->})$ satisfies ${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_{(0,\mathrm{\infty <!--\; \infty \; -->})}\mathrm{\mu <!--\; \mu \; -->}(\{f>t\}dm(t))$
At first I thought I should use $\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}$,$\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}$ and show that if $\int <!--\; \int \; -->\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}>\mathrm{\mu <!--\; \mu \; -->}(X)\log <!--\; \; -->(\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)})$, then
$\int <!--\; \int \; -->\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\mathrm{\mu <!--\; \mu \; -->}(\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\})\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}+{\int <!--\; \int \; -->}_{\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}}\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}$ and therefore
$\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}(\mathrm{\mu <!--\; \mu \; -->}(X)-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->}(\{|f|<\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}))=\log <!--\; \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}(\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(X)}\}))<{\int <!--\; \int \; -->}_{\mathrm{\mu <!--\; \mu \; -->}(\{|f|\ge <!--\; \ge \; -->\})}\log <!--\; \; -->fd\mathrm{\mu <!--\; \mu \; -->}$
which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?

A measure $\mu :B_{\mathbb{R}}\to [0,\infty ]$ is locally finite if $\mu (K)<\infty$ for all K compact. Show that any locally finite measure is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite. Give an example of a $\mathrm{\sigma <!--\; \sigma \; -->}$-finite measure on $\mathbb{R}$, which is not locally finite.
Attempt: Since $\mathbb{R}=\underset{n\in <!--\; \in \; -->\mathbb{Z}}{\bigcup <!--\; \bigcup \; -->}[n,n+1]$ and each $[n,n+1]$ is compact, we have $\mathrm{\mu <!--\; \mu \; -->}([n,n+1])<\mathrm{\infty <!--\; \infty \; -->}$ for each $n$ and $\mathrm{\mu <!--\; \mu \; -->}$ is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite. Is this part of the problem correct?
For the second part, I have no idea. I wanted to take counting measure on a collection of subsets of $\mathbb{R}$, but I know this measure is locally finite on the integers and not so with the usual topology. But I know this measure is not sigma finite on $\mathbb{R}$. Any suggestions?

I have the following proof in my notes:
$(C[0,1],\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_1)$ is not complete
Consider the sequence of functions
$f_n(t)=\{\begin{array}{lll}\frac{1}{\sqrt{t}}& ,& 1/n\le <!--\; \le \; -->t\le <!--\; \le \; -->1\\ \sqrt{n}& ,& 0\le <!--\; \le \; -->t\le <!--\; \le \; -->\frac{1}{n}\end{array}$
It can easily be shown that it is Cauchy in $\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_1$ So we have that $\mathrm{\forall <!--\; \forall \; -->}\mathrm{\epsilon <!--\; \epsilon \; -->}\mathrm{\exists <!--\; \exists \; -->}{N}_{\mathrm{\epsilon <!--\; \epsilon \; -->}}$ such that $\Vert <!--\; \Vert \; -->f_n-<!--\; -\; -->f{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}\le <!--\; \le \; -->\mathrm{\epsilon <!--\; \epsilon \; -->}\mathrm{\forall <!--\; \forall \; -->}m,n>N_{\mathrm{\epsilon <!--\; \epsilon \; -->}}$
Suppose $f_n$ converges, Then $\mathrm{\exists <!--\; \exists \; -->}f\in <!--\; \in \; -->C[0,1]$ such that $\Vert <!--\; \Vert \; -->f_n-<!--\; -\; -->f{\Vert <!--\; \Vert \; -->}_1\to <!--\; \to \; -->0.$
Then $f_{n_k}\to <!--\; \to \; -->f\text{ a.e in}[0,1]$
$f(t)=\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{f}_{n_k}(t)=\frac{1}{\sqrt{t}}\text{ a.e in}[0,1]$. Absurd.
I'm just failing to understand the last line. Can someone please explain
1. what is the absurd?
2. And how does pointwise convergence follows from a.e convergence (last line). Am I supposed to understand this $f_{n_k}\to <!--\; \to \; -->f\text{ a.e in}[0,1]$ as convergence in the real numbers, with the absolute value for every fixed t ($|f_{n_k}(t)-<!--\; -\; -->f(t)|\text{ a.e in}[0,1]$) ?

Define the operator $Kf(x)={\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy.$ by
$Kf(x)={\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy.$
where $k(x,y)$ is a continuous complex valued function on the unit square. $[0,1]\times <!--\; \times \; -->[0,1]$ Also assume that
$su{p}_{x\in <!--\; \in \; -->[0,1]}{\int <!--\; \int \; -->}_0^1|k(x,y)|dy<1$
How to show that given $g\in <!--\; \in \; -->(C[0,1],\mathbb{C})$, the integral equation
$f(x)=g(x)+{\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy$
has exactly one continuous solution $f\in <!--\; \in \; -->(C[0,1],\mathbb{C})$?
Note: I only know introductory measure theory and functional analysis, so I am suppose to do this problem without heavy integral function theory.
One way to get started is to show that K is compact and its operator norm $||K||<1$ because we are given that $su{p}_{x\in <!--\; \in \; -->[0,1]}{\int <!--\; \int \; -->}_0^1|k(x,y)|dy<1$ Then we know that $(I-<!--\; -\; -->K)$ is invertible and its inverse is given by the Neumann series. At this point, I'm not sure how to proceed.

I am reading Rudin's RCA and have a question regarding the proof of Jensen's inequality.
Let $\mathrm{\mu <!--\; \mu \; -->}$ be a positive measure on a $\mathrm{\sigma <!--\; \sigma \; -->}$-algbebra $\mathfrak{M}$ in a set $\mathrm{\Omega <!--\; \Omega \; -->}$, so that $\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\Omega <!--\; \Omega \; -->})=1$. If $f$ is a real function in $L^1(\mathrm{\mu <!--\; \mu \; -->})$, if $a<f(x)<b$ for all $x\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$, and if $\mathrm{\phi <!--\; \phi \; -->}$ is convex on $(a,b)$ then
$\mathrm{\phi <!--\; \phi \; -->}({\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\text{d}\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}.$
After a few steps Rudin obtains the following inequality:
$\mathrm{\phi <!--\; \phi \; -->}(f(x))-<!--\; -\; -->\mathrm{\phi <!--\; \phi \; -->}(t)-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}(f(x)-<!--\; -\; -->t)\ge <!--\; \ge \; -->0$
for every $x\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$. Here, $t={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\text{d}\mathrm{\mu <!--\; \mu \; -->}\in <!--\; \in \; -->(a,b)$ and $\mathrm{\beta <!--\; \beta \; -->}$ is a real number. If $\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f\in <!--\; \in \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$ then the inequality can be obtained by integrating both sides. However, I am not sure how to proceed in the case $\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f\notin <!--\; \notin \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$. Rudin states that in this case, the integral ${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}$ is defined in the extended sense
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{+}\text{d}\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{-<!--\; -\; -->}\text{d}\mathrm{\mu <!--\; \mu \; -->}$
where
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{+}\text{d}\mathrm{\mu <!--\; \mu \; -->}=\mathrm{\infty <!--\; \infty \; -->}\text{ and }{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{-<!--\; -\; -->}\text{d}\mathrm{\mu <!--\; \mu \; -->}\text{ is finite.}$
I am unable to show the existence of ${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}$ as defined above.
Any help would be greatly appreciated. Thank you.

How to find $li{m}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}su{p}_{x>0}{1}_{(0,t_n)}(x)$. It is given that $t_n\ge <!--\; \ge \; -->0,t_n\to <!--\; \to \; -->0$ as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ and 1 denotes indicator function. It seems that limit should be simply zero but how to justify it? I'm facing this problem while trying to prove a result in banach spaces.

If I define the integral for ${\int <!--\; \int \; -->}_a^b\mathbf{\text{F}}(t)dt$ as $({\int <!--\; \int \; -->}_a^b{F}_1(t)dt,{\int <!--\; \int \; -->}_a^b{F}_2(t)dt,\dots <!--\; \dots \; -->,{\int <!--\; \int \; -->}_a^b{F}_n(t)dt)$. How do I then show that
$|{\int <!--\; \int \; -->}_a^b\mathbf{\text{F}}(t)dt|\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^b|\mathbf{\text{F}}(t)|dt?$
If we write out the inequality we have
$\sqrt{{({\int <!--\; \int \; -->}_a^b{F}_1(t)dt)}^2+{({\int <!--\; \int \; -->}_a^b{F}_2(t)dt)}^2+\cdots <!--\; \cdots \; -->+{({\int <!--\; \int \; -->}_a^b{F}_n(t)dt)}^2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^b\sqrt{F_1(t{)}^2+F_2(t{)}^2+\cdots <!--\; \cdots \; -->+F_n(t{)}^2}dt.$
attempt:
If I use Jenssen's inequality I get for the left side:
$$\begin{gathered} \sqrt{{({\int <!--\; \int \; -->}_a^b{F}_1(t)dt)}^2+{({\int <!--\; \int \; -->}_a^b{F}_2(t)dt)}^2+\cdots <!--\; \cdots \; -->+{({\int <!--\; \int \; -->}_a^b{F}_n(t)dt)}^2}\le <!--\; \le \; -->\sqrt{{\int <!--\; \int \; -->}_a^b{F}_1^2(t)dt+{\int <!--\; \int \; -->}_a^b{F}_2^2(t)dt+\cdots <!--\; \cdots \; -->+{\int <!--\; \int \; -->}_a^b{F}_n^2(t)dt} \\ =\sqrt{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_a^b{F}_i^2(t)dt}=\sqrt{{\int <!--\; \int \; -->}_a^b\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{F}_i^2(t)dt}. \end{gathered}$$
The problem I have now is that the square root function is concave, so if I use Jenssen again, I get the opposite inequality as the one I need. Any idea on how to solve this?

State the number of possible triangles that can be formed using the given measurements, then sketch and solve the triangles, if possible.
$m\mathrm{\angle <!--\; \angle \; -->}A={48}^{\circ <!--\; \circ \; -->},c=29,a=4$

Suppose you are living on a straight line. There are several satellites at height 20,000 kilometers and you get readings saying that satellite 1 is directly above the point $x_1\pm {10}^{-10}$ and is at a distance $h_1=21,000\pm {10}^{-2}$ from you, satellite 2 is directly above $x_2\pm {10}^{-10}$ and at a distance $h_2=52,000\pm {10}^{-2}$. Where are you ($x_0$) and to what accuracy? Hint: Consider separately the cases $x_1<x_2$ and $x_2>x_1$.
My results:
Actually we have two equations:
$(x_1-<!--\; -\; -->x_0{)}^2+(2\cdot <!--\; \cdot \; -->{10}^4{)}^2=h_1^2$
$(x_2-<!--\; -\; -->x_0{)}^2+(2\cdot <!--\; \cdot \; -->{10}^4{)}^2=h_2^2$
We can express $x_0$ from them as:
$x_0=\frac{x_1+x_2}{2}+\frac{(h_1-<!--\; -\; -->h_2)(h_1+h_2)}{2(x_2-<!--\; -\; -->x_1)}$
But how to compute accuracy of measurement? I'm troubling with $x_2-<!--\; -\; -->x_1$ in denominator.