# Math Help - Need to Prove Almost Everywhere Result

1. ## Need to Prove Almost Everywhere Result

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.

2. Originally Posted by measureman
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.
The set $\{x\in\mathbb{R}:f(x) = g(x)\}$ is dense (because its complement has measure 0) and closed (because it is the inverse image of 0 under the continuous function $f-g$). So it must be the whole of $\mathbb{R}.$

Originally Posted by measureman
Let f : [a, b] -> R be a strictly increasing function on [a, b]. Then f is Lebesgue measurable.
For each real number $\alpha$, the set $\{x\in\mathbb{R}:f(x)<\alpha\}$ is an interval and is therefore measurable.

3. 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.

4. Originally Posted by measureman
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.
It really is that simple, and the same argument shows that you don't need the function to be strictly increasing. Weakly increasing will do just as well.

5. Originally Posted by Opalg
It really is that simple, and the same argument shows that you don't need the function to be strictly increasing. Weakly increasing will do just as well.
Thanks

How does having a complement if zero measure imply being dense? Don't we have to use the fact that f and g are continuous real valued functions?

6. 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 $\lambda$-almost everywhere.

If f and g are not equal everywhere then the is a point $x_0$ with $f(x_0)\ne g(x_0)$. Use the continuity of f–g to show that there is an open neighbourhood of $x_0$ with $f(x)\ne g(x)$ throughout that neighbourhood. Thus the set on which f and g differ contains a nonempty open set and so cannot have Lebesgue measure zero.