For a proof: first consider the set B of the numbers x in [0,1] with no "9" in their decimal expansion. B is a subset of the set of the numbers in [0,1] with no "9" among their first decimals. However, we have (where is Lebesgue measure), hence .
Then is obtained from in a "smooth" way (taking the square root): . Here is one way to show that this implies : (the function we integrate is 0 almost everywhere).
There are more general conditions that ensure that the image of a negligible set is negligible (for instance you can prove this is true if the function is Lipschitz).
Notice I made a few corrections; notably, I guessed the assumption actually is (otherwise the question is in the assumption).2.
Let be a measured space and a measurable function.
Say why belongs to .
We assume that
Show that there exists such that
For the first question, your idea is right, just notice that means for some closed (hence measurable) subset . So that .
For the second question, notice is the union of the (increasing) sequence of sets , ...
Au plaisir ,