# Meaning of measure zero

My book describes measure zero as following:

A set of points on the $x$-axis is said to have measure zero if the sum of the lengths of intervals enclosing all the points can be made arbitrarily small. If $f(x)$ is bounded in $[a,b]$, then a necessary and sufficient condition for the existence of $\int_a ^b f(x)dx$ is that the set of discontinuities have measure zero.

I don't really understand what is said here. How can the sum of the lengths of the intervals be made arbitrarily small? Wouldn't you have to move the points together? The only way I would see a set of points not obeying this is if it contains an infinite amount of points. Can someone clarify the meaning of 'measure zero' and its relation to the integral?

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you observe correctly that a set of finitely many points always have measure 0. Consequently if $f$ has finitely many discontinuities then it is Riemann integrable. – mez Feb 24 '13 at 12:59
In real analysis, this 'measure' means Lebesgue measure. A function is Riemann integrable iff the Lebesgre measure of the set of discontinuities is zero, which means this function is 'almost continuous everywhere'. – Chao Chen Feb 24 '13 at 13:04

The book's description is a bit unclear. Suppose we are given our collection of points $S$ on the real line. What it means is that, for any number $\epsilon>0$, we can choose a collection of intervals $I_1=(a_1,b_1)$, $I_2=(a_2,b_2)$, $\ldots$ such that $S$ is contained in the union $\bigcup_{k=1}^\infty I_k$ of all the intervals, and such that the sum of the lengths of all the intervals $$\sum_{k=1}^\infty m(I_k)=\sum_{k=1}^\infty (b_k-a_k),$$ is less than or equal to $\epsilon$. Since $\epsilon$ can be any number we want, this is the sense in which the sum of the lengths of the intervals can be made arbitrarily small.

The concept of measure is based on our intuition that we can assign some sets to have sizes. Of course, these sizes should behave in reasonable ways (if a set $T$ is contained in a set $S$, then the size of $T$ should be less than or equal to the size of $S$; if $S$ and $T$ have no points in common, the size of their union should be the sum of their individual sizes; etc.) Measure 0, in this context, involves a technical definition of "size" of some subsets of the real line, called the Lebesgue measure. There is a branch of mathematics called measure theory where the details of how different "size" functions can work are explored.

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not the union of all the intervals, but the intersection . . . right? each interval has to enclose all the points. – dbliss Oct 29 '15 at 21:53
@dbliss: No, it is not true that each interval must contain all the points, just that each point is in at least one interval. – Zev Chonoles Oct 29 '15 at 22:07

I think the best way to answer this is by giving an example. If we look at the set $X = \mathbb{Q} \cap [0,1]$. Which is countable since it's a subset of $\mathbb{Q}$ which is countable (if you aren't familiar with the word "countable" it really just means that you can make a list with a first element a second element, .... an n'th element etc such that all elements in the set are on this list, the integers and rational numbers are countable, but the reals aren't). Now So $\mathbb{X}= \{q_i\}_{i}$ for rational numbers $q_i$ in this set. We have given an $\epsilon > 0$ We let $I_1 = (q_1 - \epsilon/4 , q_1 + \epsilon/4)$ and let $I_n = (q_n - \epsilon/(2^{n+1}), q_n + \epsilon/(2^{n+1}))$ etc. Notice that $\cup_{n} I_n$ contains $X$ and this is what the author means by "intervals enclosing all the points". Also notice that the length of the n'th interval $I_n$ is $\epsilon/(2^n)$ hence the sum of all these intervals is given by $\sum_{n = 1}^{\infty} \epsilon/(2^n) = \epsilon$ and since $\epsilon$ was arbitrary it follows that the sum of the length of intervals enclosing all the points of $X$ can be made arbitrary small. To see what this has to do with the existence of the integral I refer you to appendix 7.9 in these notes http://www.mat.univie.ac.at/~gerald/ftp/book-fa/fa.pdf . You might have to read some of the earlier sections in chapter 7 to understand this, and you shouldn't feel down if you don't understand it all as it is quite sloppily written in my opinion.

Added: I will try to give an intuitive explanation of what this has to do with the integral. We know that the integral $\int_{a}^{b}f(x) \; dx$ exists if the function is continous. THat is that the upper and lower riemann integrals approach the same finite limit. Now what if we for example changed the value of f at finitely many points say $a_1, \ldots a_n \in [a,b]$ and made the function discontinous at these points, but let it stay the same at all other points. Could this possibly make the lower Riemann integral different from the upper? The answer is no. This is basically because the lower riemann integral is given by $sup_{P} \ ;L(P,f)$ where $L(P,f) = \sum_{j=1}^{n} m_j (x_j-x_{j-1})$ and P is a partition of the interval of the form $\{a=x_0, x_1, \ldots x_n = b\}$ and $m_j = inf ; f(x), \; x\in [x_{j-1},x_j]$. Now this sum is not going to change when we change $f$at finitely many points since this sum increases(or stays the same) whenever we increase the nuber of points in the partition, hence we can just keep adding more and more points to the partitions and at each discontinuity $a_i$ the term of the sum $m_j (x_j - x_{j-1})$ where $a_j \in [x_{j-1},x_j]$ gets smaller and smaller and converges to zero, hence we can just ignore it, while at all the other points everything is as before, hence the lower integral won't change, a similar argument shows that the upper sum won't change, thus we get exactly the same integral as before. Now this result can be generalised by using very similar arguments to the following: Let $X \subseteq [a,b]$ have measure zero. we define $\bar{f}$ to take the same values as $f$ on $[a,b] \setminus X$ but $\bar{f}$ and $f$ take different values on $X$. Then the inegral $\int_{a}^{b} \bar{f} \; dx$ exists and is equal to $\int_{a}^{b} f(x) \; dx$.

I hope this clarifies what the author is talking about and makes it possible for you to follow the rest of the book.

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This answer made the most sense to me as someone who happened upon this question, for what that's worth. – Moderat Mar 1 '13 at 18:35
I don't know what you mean when you say the text is sloppily written. To me it looks like the author have spent some time on the text: the topics are well chosen, the theory is balanced with examples, there are illustrating pictures and exercises. – AD. Mar 2 '13 at 7:09
What I mean by sloppily written is that the author skips a lot of step in his proofs. Furthermore quite a few of the propositions have either more assumptons than needed or less. this is also the case in several of the excercises. It seems that the author has aimed this text at applied mathematicians who might not be too interested in the proofs, but who want to get an introducion to functonal analysis and measure theory and see some applications of it. The only reason I linked it is that it's the only free book I know about which links the lebesgue integral with the riemann integral. – Oliver E. Anderson Mar 2 '13 at 9:07
Thanks -JJR. If you wish to study measure theory, I can recommend the book "The Elements of Integration and Lebesgue Measure" by Robert G. Bartle which is one of my favorites on this subject. The first part develops measure theory in it's general form, while the second part focuses exclusively on the Lebesgue measure which is the one your author is talking about really. According to the author you can read part two before reading part 1 if you want to. – Oliver E. Anderson Mar 2 '13 at 9:15
@VikrantDesai The rational numbers are countable which means we can order them with a first element, a second element , etc (more formaly for each $x \in \Bbb Q$ we can find exactly one element $n(x) \in \mathbb{N}$ corresponding to $x$. Then for each $x \in \Bbb Q$ Let $U_{n(x)}$ be the box $[x-\epsilon/2^{n(x)+1}, x+ \epsilon/2^{n(x)+1}]$, then $v(U_{n(x)}) = \epsilon/2^{n(x)}$. Then we have that $\sum_{i =1}^{\infty} v(U_i) = \sum_{x \in \Bbb{Q}} \epsilon/2^{n(x)} = \sum_{j=1}^{\infty} \epsilon/2^j = \epsilon (\sum_{j=1}^{\infty} 1/2^j) = \epsilon \cdot 1 = \epsilon.$ – Oliver E. Anderson Jul 12 '15 at 22:03

Your definition, as stated, is not quite complete. The problem has to do with how we might evaluate the sum of the lengths of uncountably many intervals.

More specifically: given any set $S \subset \mathbb R$,we can express it as the union of the intervals $\{[x,x] : x \in S\}$, each of which has length 0. So if we allow that the sum of the lengths of uncountably many zeroes is zero, then we get that all subsets of $\mathbb R$ have measure zero. Which is not what we want!

So you have to be more careful: a set $S \subset \mathbb R$ has measure zero if the sum of the lengths of a countable collection of intervals enclosing the points can be made arbitrarily small. This makes sense, because the sum of a countable set of non-negative real numbers is either $\infty$ or some well-defined real number.

When I say that the sum of a collection of intervals can be made arbitrarily small, I mean that if you give me any $\epsilon > 0$, then I can show you a countable collection of intervals enclosing each point of $S$, whose total length is $\le \epsilon$.

Using the argument in the second paragraph, this gives us immediately that any countable subset of $\mathbb R$ has measure zero. But there are also uncountable subsets of $\mathbb R$ with measure zero. The canonical example is the Cantor set.

This is a fascinating subject! If you don't know what a countable set is, you're lucky: just look up "Countable set" in Wikipedia, and you will find a whole new world opening up before you.

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I don't really understand it yet, What do they mean with 'enclosing the points'? – Ylyk Coitus Feb 25 '13 at 14:23
The way I currently understand it is: Measure zero means that if you have a set of points with measure zero, it means that you can add an infinitely small line segment (interval) to each side of the point (thus enclosing it), so the sum of the line segments enclosing all of the points of a set can be made arbitrarily small. – Ylyk Coitus Feb 25 '13 at 14:35
I however don't see why this could be of any use. What I would understand, is that would be important to know that you could enclose a certain point with arbitrarily big line segments, since you know that you have a continuous function if this is true. – Ylyk Coitus Feb 25 '13 at 14:36
@YlykCoitus I did not get the last sentence, what do you mean? – AD. Mar 2 '13 at 7:11
@YlykCoitus Also, is this post about understanding zero sets or about understanding why these are interesting? – AD. Mar 2 '13 at 7:13

Okay, a bunch of things. First something (call it $x$) being arbitrarily small means that you give me any (positive) $\epsilon$, no matter how small, $x$ will be less than your $\epsilon$.

Second, the paragraph says that the SUM of the length of the intervals containing the points can be made arbitrarily small. It doesn't have to be a SINGLE interval. If we were working with only a SINGLE interval then yes, the points would have to be moved together to make the interval arbitrarily small. But here we do allow several intervals if necessary so we can make each one of them arbitrary small meaning their sum can be made arbitrarily small.

Here is an easy example. Every finite set has measure zero. Consider the set $\{0,1\}$ for example. This set has measure zero because no matter how small of an $\epsilon$ you give me, I can find you two small (non-overlapping) intervals, one containing zero and the other containing one, where the sum of the length of two intervals will be less than $\epsilon$. If you give me $\epsilon=0.0001$ then I will tell you

$$[-0.00001,0.00001] \textrm{ and }[0.99999,1.00001].$$

Both of these intervals have length 0.00002 so their SUM is 0.00004 which is less than you $\epsilon=0.0001$. This is easy to see for finite number of points but it turns out that even for some infinite sets, this is also possible. For example, if we consider all of the unit fractions $\{1,1/2,1/3,1/4,1/5,...\}$ between zero and one, then this set also has measure zero. I can give a list of small enough intervals around each fraction such that the sum of all their lengths will be less than your $\epsilon$. In this case we will have infinite number of intervals but the sum of their lengths will be finite just like how the infinite sum

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=2$$

can have a finite value.

Measure here means Lebesgue Measure (there are others) and measure generalizes the "length" or "size" of a set. It allows us to measure sets which we couldn't do otherwise. We still can't measure EVERY set out there on the real number line but we can measure a lot more than we could before. So a set having measure zero means that the set is somehow "really small". It isn't too big. If it is too big then its measure will be a positive number.

The relation to the Riemann integral is that if $f(x)$ doesn't have "too many" discontinuities on its domain of integration, then it is integrable. The exact criteria is that the set of discontinuities must have measure zero. This was a big question for a while actually, that exactly what are the conditions for a function to be (Riemann) integrable. The integral is just the (signed) area under the curve and clearly the function doesn't need to continuous to be integrable. One discontinuity is okay. Two are okay. Three are okay and in fact certain infinite number of discontinuities (such as on the unit fractions) are also okay. But some functions with "too many" discontinuities aren't integrable. So exactly where do you draw the line? Lebesgue came along and answered this question brilliantly and developed a whole new branch of math called Measure Theory usually not touched upon in detail until graduate mathematics.

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In your first sentence, "$x$ will be less than your $\epsilon$" makes no sense. It should be something like "$x$ can be made less than your $\epsilon$". – TonyK Mar 2 '13 at 14:54
I think it would clarify more to add an explanation of why the irrational numbers between $0$ and $1$ are not measure zero in contrast to the rationals. – Todd Wilcox Mar 7 '13 at 19:21

A set of points on the x -axis is said to have measure zero if the sum of the lengths of intervals enclosing each of the points can be made arbitrarily small. If f(x) is bounded in [a,b] , then a necessary and sufficient condition for the existence of ∫ b a f(x)dx is that the set of discontinuities have measure zero.

Great clarification results by replacing "all" with "each of".

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