# Show a set is of measure zero

1. Jan 22, 2013

### stripes

I am trying to prove if f and g are Riemann integrable, then fg is also Riemann integrable using Lebesgue's integrability criterion. I already proved that a Riemann integrable function is bounded. Not much harder to show fg is too bounded. How do I show that [a,b] is of measure zero? I can't figure outa sequence whose infinite union contains [a,b] but also whose sum of lengths is less than all epsilon greater than zero. How do a construct such a sequence? Or should I go about it differently?

2. Jan 22, 2013

### kevinferreira

You mean [a,b] in $\mathbb{R}$? It doesn't have measure zero...

3. Jan 22, 2013

### Dick

What does the Lebesgue criterion tell you about the sets of discontinuities of f and g? How might those be related to the set of discontinuities of fg?

4. Jan 22, 2013

### stripes

Okay, I think I misread the theorem. The set of discontinuities must have measure zero for the function to be Riemann integrable. Now, I know the set of discontinuities of f and g are both of measure zero. What does that tell us about the set of discontinuities of fg?

Is the set of discontinuities of fg simply the union of the sets of discontinuities for f and g? If so, is the union of two sets whose measures are both zero, also zero? If any of these two statements are true, how would I go about showing that?

5. Jan 22, 2013

### jbunniii

You probably know this result: if $f$ and $g$ are both continuous at a point $x$, then $fg$ is also continuous at $x$. (If you don't know it, you should prove it; it's pretty easy.) So what does that say about the set of discontinuities of $fg$?

6. Jan 22, 2013

### stripes

Yes, it isn't hard to show fg is continuous in that case. Maybe I'm not understanding the problem completely, though. A function need not be continuous to be Riemann integrable. So then a function can be discontinuous and still have a set of discontinuities that is of measure zero. An example would be if this set is countable, like if it has one discontinuity. Therefore, I cannot assume that f and g are continuous to begin with. We could have an infinite number of discontinuities, but as long as these discontinuities are countable; i.e., have a one-to-one correspondence with the natural numbers, then the function will be Riemann integrable as long as it is bounded. Is this not correct?

7. Jan 22, 2013

### jbunniii

You asked whether the set of discontinuities of $fg$ is the same as the union of the sets of discontinuities of $f$ and $g$. So take the result I stated: in order for $fg$ to be discontinuous at a point $x$, either $f$ or $g$ must also be discontinuous at that point. What does that tell you?

8. Jan 22, 2013

### stripes

Your statement further supports my idea that it is the union of the two sets. So fg is discontinuous at c if and only if g or f is discontinuous at c. Because a set, by definition, doesn't count the same element twice, if we are given a point of discontinuity of fg, we don't care if it's in one set, the other set, or both. So it must be the union of the two sets.

Subadditivity--so the measure of the union of two sets is equal to or less than the sum of the measures of the two sets individually? Since a measure is always positive (I think), 0 + 0 = 0, so its union has measure 0 as well.

9. Jan 22, 2013

### Dick

Define h(x)=1 if x>=0 and -1 if x<0. Pick f=h and g=h. What the discontinuities of fg? I don't think it's the exactly the union. But yes, the union has measure zero.

Last edited: Jan 22, 2013
10. Jan 22, 2013

### stripes

Is it just the same set then? While either f or g can be discontinuous for fg to be discontinuous, they end up being the same point anyways. If it is neither the union nor either set itself, then I am lost.

Your example doesn't suggest it isn't the union. The discontinuities of f is x=0, the discontinuities of g is x=0, and the discontinuities of fg is x=0. The union of {0} and {0} is {0}.

11. Jan 22, 2013

### Dick

fg is h^2=1. It's continuous everywhere. If f is discontinuous and g is discontinuous at a point x, that doesn't mean fg is discontinuous at x. It's not even true that if f is continuous and g is discontinuous then fg is discontinuous. What is true is that if f and g are continuous then fg is continuous. That's about all.

12. Jan 22, 2013

### jbunniii

Indeed. Simple counterexample: f(x) = 0 for all x.

13. Jan 22, 2013

### stripes

Okay, so the contrapositive tells us that if fg is not continuous then f or g is not continuous. So is the set of discontinuities of fg not a subset of the union of discontinuities? Because the set of all discontinuities of fg will definitely be in the set for f or g. Might be both, might be only one, but it is in at least one. So if I know for sure that the union of two sets whose measure is zero is also zero, then I can be 100% certain that any subset of this union is also of measure zero. So if this set that I am so desperately trying to find is a subset of the union, and the union's measure is also zero, then I'm in business.

14. Jan 22, 2013

### Dick

You're in business. It must be a subset of the union of the discontinuities.

15. Jan 22, 2013

### stripes

Thanks everyone. Your help is appreciated.