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I want to show that if $a\lt b$, then $(a,b)$ is not of measure zero.

My idea was to show that any interval covering $(a,b)$ is such that the sum of the lengths of the intervals is always greater than $b-a$.
So I tried induction. Suppose $n=1$, where $n$ is the number of intervals covering $(a,b)$. Then the interval consists of a single set of the form, say, $(c,d)$ such that $(a,b)\subset (c,d)$. If this is the case then obviously, we would have $d-c \gt b-a$, where I have assumed that $c\lt a$ and $b \lt d$.

Now my questions: I am a little bit lost as to how to show it for $n+1$.

Am I even approaching it correctly?

Thanks for your help.

  • 2
    @Austin: Yes, you start with the usual length on intervals and then you associate outer measure. The tricky bit is to show that you can't cover an interval $(a,b)$ by a family of intervals $I_n$ such that $\sum |I_n| \lt b - a$. See also [this MO post by Gowers](http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove/51555#51555)2011-09-09
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    @Theo I agree. I mostly wanted to match up terminology. In any event it is good to mention these things.2011-09-09
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    @All: It is Lebesgue measure on $\mathbb{R}$2011-09-09

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