The temperature in Hillsville was ${20}^{\circ}$ Celsius. What is the equivalent of this temperature in degrees Fahrenheit?

ghulamu51
2022-09-14
Answered

The temperature in Hillsville was ${20}^{\circ}$ Celsius. What is the equivalent of this temperature in degrees Fahrenheit?

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asked 2022-07-15

Properly showing that uncountable sum of measures finite nonzero sets is infinite, given they are subsets of $X$ of finite measure.

In a question where $(X,\sum ,\mu )$ is a finite measure space, I am asked to show that the set $\{x|\mu (\{x\}>0)\}$ is countable at most.

While intuitively this has to be true, because as long as it is countable any series of nonzero measures of sets must converge and this can happen for geometric sequences of measures, but for an uncountable sum, this "can't" be the case, but how to show that- I can't seem to understand.

Firstly, is this the way to solve it? Trying to reach a contradiction, or maybe uncountable sum of elements is generally meaningless as an expression? I'm not sure, what

In a question where $(X,\sum ,\mu )$ is a finite measure space, I am asked to show that the set $\{x|\mu (\{x\}>0)\}$ is countable at most.

While intuitively this has to be true, because as long as it is countable any series of nonzero measures of sets must converge and this can happen for geometric sequences of measures, but for an uncountable sum, this "can't" be the case, but how to show that- I can't seem to understand.

Firstly, is this the way to solve it? Trying to reach a contradiction, or maybe uncountable sum of elements is generally meaningless as an expression? I'm not sure, what

asked 2022-05-21

We know when $(X,S,\mu )$ is a measure space, ${M}_{\mathbb{R}}(S)$ consists signed (or real) measures on $(X,S)$ is Banach space. The set V consists of the measures satisfying $dv=hd\mu $ for $h\in {L}^{1}(\mu )$ is closed but how we can show that when $\mu $ is Lebesgue measure and S would be the cllections of Borel sets, the set V is not separable.

asked 2022-06-22

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by

$f(x)=\{\begin{array}{ll}x[1+\mathrm{sin}(\mathrm{ln}x)]& \text{if}x0\\ 0& \text{if}x=0\\ x+\sqrt{-x{\mathrm{sin}}^{2}(\mathrm{ln}|x|)}& \text{if}x0.\end{array}$

Find the values of the four Dini derivatives of $f$ at $x=0.$

I could easily find out the values of the upper right ${D}^{+}f$ and lower right ${D}_{+}f$ Dini derivatives at $x=0,$ values being 2 and 0 respectively. But, in case of upper left Dini derivative of $f$ at $x=0,$ we have:

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to 0-}{lim\u2006sup}\frac{f(h)-f(0)}{h}\\ \\ & =1+\underset{h\to 0-}{lim\u2006sup}\frac{\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}|h|)}}{h}.\end{array}$

Next putting: $-h=p$ so that $p\to 0+,$, I obtained:

${D}^{-}f(0)=1-\underset{p\to 0+}{lim\u2006inf}\frac{|\mathrm{sin}(\mathrm{ln}|p|)|}{\sqrt{p}}.$

Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term $\sqrt{p}$ present in the denominator. I am confused. Please give some insights. Thanks in advance.

$f(x)=\{\begin{array}{ll}x[1+\mathrm{sin}(\mathrm{ln}x)]& \text{if}x0\\ 0& \text{if}x=0\\ x+\sqrt{-x{\mathrm{sin}}^{2}(\mathrm{ln}|x|)}& \text{if}x0.\end{array}$

Find the values of the four Dini derivatives of $f$ at $x=0.$

I could easily find out the values of the upper right ${D}^{+}f$ and lower right ${D}_{+}f$ Dini derivatives at $x=0,$ values being 2 and 0 respectively. But, in case of upper left Dini derivative of $f$ at $x=0,$ we have:

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to 0-}{lim\u2006sup}\frac{f(h)-f(0)}{h}\\ \\ & =1+\underset{h\to 0-}{lim\u2006sup}\frac{\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}|h|)}}{h}.\end{array}$

Next putting: $-h=p$ so that $p\to 0+,$, I obtained:

${D}^{-}f(0)=1-\underset{p\to 0+}{lim\u2006inf}\frac{|\mathrm{sin}(\mathrm{ln}|p|)|}{\sqrt{p}}.$

Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term $\sqrt{p}$ present in the denominator. I am confused. Please give some insights. Thanks in advance.

asked 2022-04-02

Tell what kind of measurement scale is appropriate for the following information.

1. Eye color

2 Gender

3. Race

4. Speed of sound

5. Grades

6. Sizes of T-shirts

Explain: Mean and Variance Discrete Variable.

1. Eye color

2 Gender

3. Race

4. Speed of sound

5. Grades

6. Sizes of T-shirts

Explain: Mean and Variance Discrete Variable.

asked 2022-05-26

For $a>0$ and a r.v. $X$, is it true that ${\int}_{0}^{\mathrm{\infty}}{a}^{p-1}\mathbb{P}[X\ge a]\mathrm{d}a=\mathbb{E}[{\int}_{0}^{X}{a}^{p-1}\mathrm{d}a]$? If so, why?

asked 2022-05-20

Let $f$ be a nonegative measurable function on $\mathbb{R}$. Show that

$\overline{){\displaystyle \underset{n\to \mathrm{\infty}}{lim}{\int}_{-n}^{n}f={\int}_{\mathbb{R}}f}}$

Of course

$\mathrm{\forall}n,{\int}_{-n}^{n}f\le {\int}_{\mathbb{R}}f$

To show the reverse, I feel that I need to use Fatou's lemma; but can't seem to get the right argument.

$\overline{){\displaystyle \underset{n\to \mathrm{\infty}}{lim}{\int}_{-n}^{n}f={\int}_{\mathbb{R}}f}}$

Of course

$\mathrm{\forall}n,{\int}_{-n}^{n}f\le {\int}_{\mathbb{R}}f$

To show the reverse, I feel that I need to use Fatou's lemma; but can't seem to get the right argument.

asked 2022-05-11

I was trying to solve the next question:

let $X$ be a finite measure space ( $\mu (X)<\mathrm{\infty}$ ) and let $f\in {L}^{1}(X,\mu )$, $f(x)\ne 0$ almost everywhere.

Show that for each measurable subset $E\subset X$:

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu =\mu (E)$

My idea for a solution is to use Fatou's lemma:

In one direction:

${\int}_{E}\underset{n\to \mathrm{\infty}}{lim\u2006inf}|f{|}^{\frac{1}{n}}d\mu \le \underset{n\to \mathrm{\infty}}{lim\u2006inf}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu $

and

${\int}_{E}\underset{n\to \mathrm{\infty}}{lim\u2006inf}|f{|}^{\frac{1}{n}}={\int}_{E}1d\mu =\mu (E)$

So we get:

$\mu (E)\le \underset{n\to \mathrm{\infty}}{lim\u2006inf}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu $

In the other direction, I thought of maybe saying that we know there is an $\epsilon >0$

And an $N\in \mathbb{N}$ so for all $n>N$ we get that $1+\epsilon >|f{|}^{\frac{1}{n}}$ which means $1+\epsilon -|f{|}^{\frac{1}{n}}>0$

and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to $\mu (E)$

Is it valid? Am I missing something?

If I do, what can I do to prove the other direction?

Thank you!

let $X$ be a finite measure space ( $\mu (X)<\mathrm{\infty}$ ) and let $f\in {L}^{1}(X,\mu )$, $f(x)\ne 0$ almost everywhere.

Show that for each measurable subset $E\subset X$:

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu =\mu (E)$

My idea for a solution is to use Fatou's lemma:

In one direction:

${\int}_{E}\underset{n\to \mathrm{\infty}}{lim\u2006inf}|f{|}^{\frac{1}{n}}d\mu \le \underset{n\to \mathrm{\infty}}{lim\u2006inf}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu $

and

${\int}_{E}\underset{n\to \mathrm{\infty}}{lim\u2006inf}|f{|}^{\frac{1}{n}}={\int}_{E}1d\mu =\mu (E)$

So we get:

$\mu (E)\le \underset{n\to \mathrm{\infty}}{lim\u2006inf}{\int}_{E}|f{|}^{\frac{1}{n}}d\mu $

In the other direction, I thought of maybe saying that we know there is an $\epsilon >0$

And an $N\in \mathbb{N}$ so for all $n>N$ we get that $1+\epsilon >|f{|}^{\frac{1}{n}}$ which means $1+\epsilon -|f{|}^{\frac{1}{n}}>0$

and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to $\mu (E)$

Is it valid? Am I missing something?

If I do, what can I do to prove the other direction?

Thank you!