# Graph of f has Measure Zero

1. Apr 10, 2008

### varygoode

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I'm pretty clueless as to what's going on here. If someone can just please lead me in the right direction, I would be quite grateful.

2. Apr 10, 2008

### Dick

Start with a really simple case. Take Q=[0,1] in R. Take f(x)=x. Then the 'graph' is the diagonal of the square [0,1]x[0,1] in R^2. Can you show that graph has zero measure in R^2? How would you modify that argument to handle the general case? Hint: f is in fact uniformly continuous since the domain is compact.

3. Apr 10, 2008

### varygoode

I think I can easily find countably many rectangles to cover the diagonal you are talking about. Something like if I take all the intervals on the line, all of length 1 let's say, then I can cover the interval by n squares with height epsilon/n. Then if I take the union of them, I'll get the volume is less than epsilon in summation.

But how do I generalize this?

4. Apr 10, 2008

### Dick

That's not super clear, but ok. So now instead of f(x)=x, take f(x) to be any continuous function. Do you believe f(x) is uniformly continuous? Can you prove it? If so then just recite the definition of uniform continuity. For every epsilon>0 there exists a delta>0 such that... Given that how many rectangles do you need to cover [0,1]? What's the area of each rectangle? What's the total area? Now let epsilon approach 0.

5. Apr 10, 2008

### varygoode

It's uniformly continuous since it is continuous on a compact set, right?

So for every epsilon > 0 there exists delta > 0 s.t. |x-y|< delta implies |f(x) - f(y)| < epsilon. I've got that I think.

But I need something else to connect that and the rectangles. I'm horrible at picturing things, so that won't help. I'm just not sure how to define the rectangles in order to ensure they cover G(f). I simply don't see it.

What is the connection?

6. Apr 10, 2008

### Dick

Draw rectangles that are delta in x by epsilon in y in size. How many do you need to cover the range of x in [0,1]? Multiply that by the area of each one. You are right on the uniform thing.

Last edited: Apr 10, 2008
7. Apr 12, 2008

### varygoode

Wait, why are we talking about [0,1]? Damn, I'm completely lost here.

Can you give me this explanation in some mathematically explicit terms? I'm having trouble following what's going on here.

8. Apr 12, 2008

### Dick

I am trying to get you to figure out how to solve an less complicated version of the problem so the notational details don't get in the way of understanding how the proof works. If you can do the problem for the case f:[0,1]->R, I think you can figure out how to generalize it to f:Q->R.