The set is dense (because its complement has measure 0) and closed (because it is the inverse image of 0 under the continuous function ). So it must be the whole of
For each real number , the set is an interval and is therefore measurable.
Hi everyone, I'm struggling to prove the following two results:
Let f, g : R -> R be continuous real-valued functions on R such that f = g
lambda-almost everywhere (f = g everywhere except on a set of Lebesgue measure 0). Then f = g everywhere.
Let f : [a, b] -> R be a strictly increasing function on [a, b]. Then f is
Lebesgue measurable.
Can someone give me a hint on where to go with this?
Thanks.
The set is dense (because its complement has measure 0) and closed (because it is the inverse image of 0 under the continuous function ). So it must be the whole of
For each real number , the set is an interval and is therefore measurable.
Don't know anything about denseness - shall have to go look it up.
For the second one, this is what I was thinking, but I thought surely it cannot be this simple. Could you give me an example of a function that is not strictly increasing, that is not Lebesgue measurable? I can't see why this is true only for strictly increasing functions.
Thanks.
Maybe you'll find it easier to use another method: prove the contrapositive. So assume that f and g are continuous functions that are not equal everywhere. You want to show that they cannot be equal -almost everywhere.
If f and g are not equal everywhere then the is a point with . Use the continuity of f–g to show that there is an open neighbourhood of with throughout that neighbourhood. Thus the set on which f and g differ contains a nonempty open set and so cannot have Lebesgue measure zero.