Elleanor Mckenzie
2021-02-06
Answered

How do you calculate the following retaining the correct number of significant figures $12.432\times 3=$ and $208\times 62.1=$

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avortarF

Answered 2021-02-07
Author has **113** answers

Step 1
Firstly we calculate the first part.
Multiply both the numbers
After that round off the result at tenth place.
$12.432\times 3=37.296$

$37.396\approx 37.4$
Step 2
Significant figures are those that have some meaning in overall value of the number.
To determine what numbers are significant and which arent,

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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!

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