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Sorry for pasting an image but I cannot write with math terminology. I have spent a day looking at this. I do not understand this proof from a book. m* stands for outer measure.enter image description here

I thought the coverings should contain the interval. How can an outer measure that makes use of coverings be smaller than the actual length of the interval. Thanks for reading.

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    The coverings in question have lengths greater $b-a+\varepsilon$ than the length $b-a$ of the interval you're working with. The outer measure equals the length of the interval. Here it's only proved that the outer measure $\leq$ the length, but $\geq$ is also true. – 2017-01-07
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    My problem is about understanding this inequality. I cannot understand why this proof is true, since the coverings of the interval should contain the interval itself. – 2017-01-07
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    The coverings do contain the interval itself. The inequality ($\leq$) holds, but in fact it's always $=$, not $<$. – 2017-01-07

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The point is that - for each $\epsilon>0$ - the set $$\mathcal{I}_\epsilon=\{]a, b+\epsilon[\}$$ is a cover of the set $$]a, b].$$ In particular, $b\in]a, b+\epsilon[$. (Do you see why?)

Now, the sum of the sizes of the open intervals in $\mathcal{I}_\epsilon$ is $b+\epsilon-a$ - so this tells us $$m^*(]a, b])\le b+\epsilon-a.$$ Since this is true for any $\epsilon>0$, we have in fact $$m^*(]a, b])\le b-a.$$