# Measure zero

1. Jun 27, 2013

### Tenshou

I don't understand what it means to have measure zero. Like I understand that the indicator functions is and what it does, but I am not exactly sure what measure zero is. Okay, I mean that I understand that:
$\int_{A} 1_{A} dx = \nu(A)$

and this is the "volume" of some region A in $\mathbb{R}^{k}$ and I understand that it has measure zero when volume is zero, but what is it useful for and what is measure zero

2. Jun 27, 2013

### Simon Bridge

You are seriously saying that you have got this far through maths education and you don't understand why zero is important or what it is?
"measure zero" - the measure of the null set - extends the concept of zero beyond arithmetic.
If you've only just been introduced to the terminology, be patient, you will soon have examples of it's use coming out your ears ;)

3. Jun 27, 2013

### Tenshou

D: But I haven't really had a formal maths education, I am all self taught :3 But do you think you could give me an example of what measure zero and explain why it has measure zero

4. Jun 27, 2013

### deluks917

Any countable subset of R (such as Q) has measure zero. However there are uncountable subsets of R with measure 0 (example the Cantor Set).

In R^n any subset that has dimension less than n has measure 0. For example the surface of a sphere as a subset of R^3. Or the graph of a continutous function in R^2.

5. Jun 27, 2013

### jbunniii

Start by considering the 1-dimensional case: measure in $\mathbb{R}^1$ is a generalization of length. For an interval of the form $[a,b]$, the length is simply $b-a$, and we define the measure of this interval to also be $b-a$. For a more general set $A \subset \mathbb{R}$, it's clearly possible to cover $A$ with countably many intervals of the form $[a,b]$, and if we add up the lengths of these intervals, then surely the "length" of $A$ can't be greater than that sum. So, we can reasonably define the "length" of $A$ to be the infimum taken over all possible ways of covering $A$ with a countable union of intervals.

Now consider some concrete examples. If $A = \{a\}$ is a set containing one point, then we may also write it as the zero-length interval $[a,a]$, so we expect its measure to be $a-a = 0$. In terms of the infimum definition, we can see that any interval of the form $[a-\epsilon, a+\epsilon]$ (where $\epsilon > 0$) will cover $A$. The length of the interval $[a-\epsilon, a+\epsilon]$ is $2\epsilon$, so the measure of $A$ is at most $2\epsilon$. This is true for any $\epsilon > 0$, so that forces $m(A) = 0$, as expected.

We can make a similar argument to show that a set containing a finite number of points also has measure zero.

Similarly, a set containing a countably infinite number of points also has measure zero. Sketch of proof: let $(x_n)$ be an enumeration of the elements of $A$. Let $\epsilon > 0$. Then we may cover each point $x_n$ with an interval of length $\epsilon / 2^n$, and the sum of the lengths of these intervals is $\sum \epsilon / 2^n = \epsilon$. Thus the measure of $A$ is at most $\epsilon$. But $\epsilon$ was arbitrary, so $m(A) = 0$.

We can even find sets with uncountably many elements, but which still have measure zero. The classic example is the Cantor set.

6. Jun 28, 2013

### Simon Bridge

Great example jbunnil :) and for completeness, could you provide an example of how the measure zero is used?

7. Jun 28, 2013

### jbunniii

Sure, there are many situations in which measure zero comes up.

One example that comes to mind is the very nice characterization of Riemann integrability: a bounded function is Riemann integrable if and only if the set of points on which it is discontinuous has measure zero.

Another even more important example is that a nonnegative Lebesgue integrable function $f$ has integral equal to zero if and only if the set of points $x$ for which $f(x) > 0$ has measure zero. This characterization allows us to use the integral to define a norm on spaces of functions.

Certainly for any integrable function, we may define $||f|| = \int |f(x)| dx$, but this is only a seminorm because we can have $||f|| = 0$ even if $f$ is not identically zero.

However, we can define an equivalence relation on the set of integrable functions, such that two functions are equivalent if they differ only on a set of measure zero. The above-defined seminorm becomes a genuine norm when we define it in the obvious way on the set of equivalence classes.

Norms are useful for many reasons, especially the fact that we can define a metric (distance) between two functions as $d(f,g) = ||f-g||$. Once we have a metric space, this gives us a nice topology, and we can now apply all the tools of topology to function spaces.

The cost of doing this is that we can no longer distinguish between functions that differ only on a set of measure zero. We have to replace the notion of pointwise convergence, for example, with convergence "almost everywhere", meaning convergence everywhere except on a set of measure zero.

But this is still a very powerful concept. For example, in general it's hard to say much about pointwise convergence of the Fourier series of functions unless they are continuous. But if we relax this to almost-everywhere convergence, we can say very profound things: if $\int |f(x)|^p dx < \infty$ for some $p > 1$, then the Fourier series of $f$ converges almost everywhere. (Proof is left to the reader. Look up the Carleson-Hunt theorem.)

8. Jun 28, 2013

### HallsofIvy

Staff Emeritus
"Fourier series" (and the Fourier integral) are very useful tools for engineers. In fact, Fourier himself was an engineer, not a mathematician. And like most engineers, he often skipped over the "details" in order to solve his problem!

It is certainly true that if we have an integrable function, satisfying the properties Fourier series required, we can find the coefficients of the Fourier series for that function by integrating. Fourier assumed that the reverse was true- that given a valid Fourier series, it converged to an integrable function which you could integrate to get those coefficients. And that is NOT always true- at least not using the Riemann integral. Since Fourier series clearly did work, it was up to the mathematicians to define a more general "integral" so that would be true.

That more general integral was the "Lebesque integral" which uses a more general definition of "measure" of a set than the "length of an interval" definition used to define the Riemann integral. In particular the function "f(x)= 1 if x is irrational, f(x)= 2 if x is rational" cannot be integrated in the Riemann sense because there is no interval that contains only rational or only irrational numbers. No matter what partition we use, the "upper sum" is always 1, the lower sum is always 0, so there is no limit process that makes them come out the same. But using the Lebesque definition of "measure" we can divide the interval, from 0 to 1, say, into precisely the rational and irrational numbers, even though those sets cannot be written as unions and intersections of invervals. The measure of the set of rational numbers from 0 to 1 is 0 (because it is countable) so the measure or the set of irrational numbers from 0 to 1 is 1- 0= 1. The (Lebesque) integral of this function from 0 to 1 is (1)(1)+ (0)(2)= 1.

One complication of this theory is that integrals cannot "distinguish" between functions that differ only on a "set of measure 0". We say that such functions are "equal almost everywhere". (My teacher for Lebesque integration was from Eastern Europe and to this day I remember him saying "almost ev-ary where".)

In modern probability theory, Lebesque integration is used extensively and we say that an event is "almost certain" if its probability is 1 except on a set of measure 0.