On proof that the outer (Lebesgue)measure of an interval is its length

Question

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,,

**

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.

**

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

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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 $x$ $\ge$ $y$ + $\epsilon$)

Answer 1

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.

Answer 2

If $x \le y+ \epsilon$ for all $ \epsilon>0$, then we can conclude that $x \le y$.

If $x>y$ then the choice $\epsilon = {1 \over 2}(x-y) >0$ leads to an immediate contradiction.

Furthermore, from the definition of $m^*$, we see that if $A \subset B$ then $m^* A \le m^* B$.