Prove that $m^*(E+x)=m^*(E)$
$begingroup$
Let $Esubsetmathbb{R}$ and $xin mathbb{R}$.
$E+x={e+x: ein E}$
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
Prove that $m^*(E+x)=m^*(E)$.
I think my proof is correct but I'd like someone to take a look.
Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.
$Esubsetbigcup_{n=1}^infty I_n$
We claim that ${I_n+x}$ covers $E+x$. To prove this, let $yin E+ximplies y=e+x$ where $ein E$. Since $Esubsetbigcup_{n=1}^infty I_n$, $ein Eimplies einbigcup_{n=1}^infty I_nimplies ein I_n$ for some $n$. So, $y=e+xin I_n+x$ for some $n$. Thus, $yinbigcup_{n=1}^infty(I_n+x)$. Hence $E+xsubsetbigcup_{n=1}^infty(I_n+x)$.
So, $m^*(E+x)lesum_{n=1}^inftyell(I_n+x)$
Since $I_n+x$ is merely the interval $I_n$ translated by $x$, $ell(I_n+x)=ell(I_n)$.
Thus, $m^*(E+x)lesum_{n=1}^inftyell(I_n)=m^*(E)$ ------------ (1)
$E$ can be written as $(E+x)+(-x)$
Using (2) with $E+x$ as $E$ and $-x$ as $x$,
$m^*(E+x+(-x))le m^*(E+x)$
$implies m^*(E)le m^*(E+x)$ ---------------- (2)
From (1) and (2),
$m^*(E+x)=m^*(E)$
QED
real-analysis proof-verification
edited Dec 2 '18 at 22:18
Winther
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
$endgroup$
add a comment |
$begingroup$
Let $Esubsetmathbb{R}$ and $xin mathbb{R}$.
$E+x={e+x: ein E}$
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
Prove that $m^*(E+x)=m^*(E)$.
I think my proof is correct but I'd like someone to take a look.
Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.
$Esubsetbigcup_{n=1}^infty I_n$
We claim that ${I_n+x}$ covers $E+x$. To prove this, let $yin E+ximplies y=e+x$ where $ein E$. Since $Esubsetbigcup_{n=1}^infty I_n$, $ein Eimplies einbigcup_{n=1}^infty I_nimplies ein I_n$ for some $n$. So, $y=e+xin I_n+x$ for some $n$. Thus, $yinbigcup_{n=1}^infty(I_n+x)$. Hence $E+xsubsetbigcup_{n=1}^infty(I_n+x)$.
So, $m^*(E+x)lesum_{n=1}^inftyell(I_n+x)$
Since $I_n+x$ is merely the interval $I_n$ translated by $x$, $ell(I_n+x)=ell(I_n)$.
Thus, $m^*(E+x)lesum_{n=1}^inftyell(I_n)=m^*(E)$ ------------ (1)
$E$ can be written as $(E+x)+(-x)$
Using (2) with $E+x$ as $E$ and $-x$ as $x$,
$m^*(E+x+(-x))le m^*(E+x)$
$implies m^*(E)le m^*(E+x)$ ---------------- (2)
From (1) and (2),
$m^*(E+x)=m^*(E)$
QED
real-analysis proof-verification
edited Dec 2 '18 at 22:18
Winther
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
$endgroup$
add a comment |
$begingroup$
Let $Esubsetmathbb{R}$ and $xin mathbb{R}$.
$E+x={e+x: ein E}$
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
Prove that $m^*(E+x)=m^*(E)$.
I think my proof is correct but I'd like someone to take a look.
Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.
$Esubsetbigcup_{n=1}^infty I_n$
We claim that ${I_n+x}$ covers $E+x$. To prove this, let $yin E+ximplies y=e+x$ where $ein E$. Since $Esubsetbigcup_{n=1}^infty I_n$, $ein Eimplies einbigcup_{n=1}^infty I_nimplies ein I_n$ for some $n$. So, $y=e+xin I_n+x$ for some $n$. Thus, $yinbigcup_{n=1}^infty(I_n+x)$. Hence $E+xsubsetbigcup_{n=1}^infty(I_n+x)$.
So, $m^*(E+x)lesum_{n=1}^inftyell(I_n+x)$
Since $I_n+x$ is merely the interval $I_n$ translated by $x$, $ell(I_n+x)=ell(I_n)$.
Thus, $m^*(E+x)lesum_{n=1}^inftyell(I_n)=m^*(E)$ ------------ (1)
$E$ can be written as $(E+x)+(-x)$
Using (2) with $E+x$ as $E$ and $-x$ as $x$,
$m^*(E+x+(-x))le m^*(E+x)$
$implies m^*(E)le m^*(E+x)$ ---------------- (2)
From (1) and (2),
$m^*(E+x)=m^*(E)$
QED
real-analysis proof-verification
edited Dec 2 '18 at 22:18
Winther
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
$endgroup$
Let $Esubsetmathbb{R}$ and $xin mathbb{R}$.
$E+x={e+x: ein E}$
Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
Prove that $m^*(E+x)=m^*(E)$.
I think my proof is correct but I'd like someone to take a look.
Let ${I_n}$ be a sequence of intervals that covers $E$ such that $m^*(E)=sum_{n=1}^inftyell(I_n)$.
$Esubsetbigcup_{n=1}^infty I_n$
We claim that ${I_n+x}$ covers $E+x$. To prove this, let $yin E+ximplies y=e+x$ where $ein E$. Since $Esubsetbigcup_{n=1}^infty I_n$, $ein Eimplies einbigcup_{n=1}^infty I_nimplies ein I_n$ for some $n$. So, $y=e+xin I_n+x$ for some $n$. Thus, $yinbigcup_{n=1}^infty(I_n+x)$. Hence $E+xsubsetbigcup_{n=1}^infty(I_n+x)$.
So, $m^*(E+x)lesum_{n=1}^inftyell(I_n+x)$
Since $I_n+x$ is merely the interval $I_n$ translated by $x$, $ell(I_n+x)=ell(I_n)$.
Thus, $m^*(E+x)lesum_{n=1}^inftyell(I_n)=m^*(E)$ ------------ (1)
$E$ can be written as $(E+x)+(-x)$
Using (2) with $E+x$ as $E$ and $-x$ as $x$,
$m^*(E+x+(-x))le m^*(E+x)$
$implies m^*(E)le m^*(E+x)$ ---------------- (2)
From (1) and (2),
$m^*(E+x)=m^*(E)$
QED
real-analysis proof-verification
real-analysis proof-verification
edited Dec 2 '18 at 22:18
Winther
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
edited Dec 2 '18 at 22:18
Winther
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
edited Dec 2 '18 at 22:18
Winther
20.6k33156
edited Dec 2 '18 at 22:18
Winther
20.6k33156
edited Dec 2 '18 at 22:18
Winther
20.6k33156
20.6k33156
asked Dec 2 '18 at 22:02
ThomasThomas
730416
asked Dec 2 '18 at 22:02
ThomasThomas
730416
asked Dec 2 '18 at 22:02
ThomasThomas
730416
730416
add a comment |
add a comment |
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