Hi,

I don't understand your last sentence. True A is uncountable, but it is negligible anyway. One fact to keep in mind: (for Lebesgue measure,) almost every real number in [0,1] haseveryfinite pattern (like "31415925") infinitely many times in its decimal expansion. In particular, almost every real number has a 9 (and even infinitely many, and arbitrarily long sequences of 9's) in its decimal expansion.

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,

Laurent.