Meaning of measure zero

Solution 1:

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.

Solution 2:

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.