Begining the proof that the outer (Lebesgue) measure of an interval is its length, i'm little confused from the start.
i know what the outer measure is and the whole process of the proof (using Heine – Borel Theorem, infimum..etc)
but, at the start of the proof, let me write,,
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it begins with the case of closed bounded interval, say, $[$a$,$b$]$. Since the open interval ($a$-$\epsilon$, $b$+$\epsilon$) contains [a,b] for each positive number $\epsilon$, we have m$^*$[a,b] $\le$ b-a +2$\epsilon$. but since this is true for each positive $\epsilon$, we must have m$^*$[a,b] $\le$ b-a.
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i cannot understand from the phrase " but since this is true…….."
how can we say m$^*$[a,b] $\le$ b-a
from m$^*$[a,b] $\le$ b-a +2$\epsilon$ ?
please give me a hint . thanks
for $x$,$y$ in $R$ and for arbitrary $\epsilon$
$x$ $\lt$ $y+$$\epsilon$ implies $x$$\lt$ $y$
can this proposition explain my question??
(that proposition is contraposition of
"$x$ $\ge$ $y$ implies there exist $\epsilon$ $\gt$ $0$ s.t $x$ $\ge$ $y$ + $\epsilon$)
Best Answer
It is true when $\epsilon=1/2$: $m^\ast[a,b]\leq b-a+2\times 1/2$
It is true when $\epsilon=1/4$: $m^\ast[a,b]\leq b-a+2\times 1/4$
It is true when $\epsilon=1/10000000$: $m^\ast[a,b]\leq b-a+2\times 1/10000000$
Since we can make $\epsilon$ arbitrarily small, it must be true that $m^\ast[a,b]\leq b-a$. If $m^\ast[a,b]$ were even a little bit bigger than $b-a$, we could find a small enough value for $\epsilon$ so that $b-a<b-a+2\epsilon<m^\ast[a,b]$, which would be a contradiction.