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So I was reading a proof for Lipschitz transformations transform measure zero sets to measure zero sets.

There the following argument using diffeomorphism $T$ is used:

"It follows by covering with cubes that there's constant $c'$ s.t. $|TI| \le c'|I|$ for any interval $I$."

What's this "covering with cubes" and why does it lead to the inequality?

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    It might refer to a use of the [Vitali covering lemma](https://en.wikipedia.org/wiki/Vitali_covering_lemma). That talks about balls, not cubes, but I think it works just as well for cubes. Hard to say for sure without a bit more contextual information, though. – 2017-02-17
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    @HaraldHanche-Olsen p. 55 in this book for example https://books.google.fi/books?id=X3d3CAAAQBAJ&pg=PA55&lpg=PA55&dq=lipschitz+measure+zero+it+follows+by+covering+cubes&source=bl&ots=7jBIeau-Qb&sig=YW5QgeY6gjwIGK5W5P4KuKtVYzE&hl=fi&sa=X&ved=0ahUKEwj6q5yaxJfSAhWGYJoKHdcGAccQ6AEIITAD#v=onepage&q=lipschitz%20measure%20zero%20it%20follows%20by%20covering%20cubes&f=false – 2017-02-17

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Okay, I think the point is as follows: Trivially, the Lipschitz map $T$ maps any set to a set whose diameter is at most $c$ times the diameter of the original. If $I$ is a cube of side $a$ in $\mathbb{R}^n$, then, $TI$ has diameter at most $\sqrt{n}ca$. And that, in its turn, can then be placed inside a cube of side $\sqrt{n}ca$, which will have measure $n^{n/2}c^na^n$. That is $n^{n/2}c^n$ times the measure of the original.

If you try the same computation with an interval, it will fail, since the interval can have one very long side and some very short ones, so the $n$th power of the diameter is a very poor estimate of the measure. The cure is to cover the interval in cubes, use the previous estimate on each cube, and conclude the measure still expands at most by a constant factor.

This covering of cubes can be done with great precision, if you wish. But for the present purpose, you can do a very rough job: Let $\delta$ be the smallest side length of $I$, and stack cubes of side $\delta$ in layers to cover $I$. The resulting stack of cubes will certainly not have a measure greater than $2^n$ times the original, and that's good enough! Any constant is good enough. (Of course, by picking smaller cubes, you can get the constant as close to $1$ as you wish.)