# Tuesday August 15th > See Folland's Real Analysis, definitely a recommended reference. Possible first day question: how can we "measure" a subset of $\RR$? We'd like bigger sets to have a higher measure, we wouldn't want removing points to increase the measure, etc. This is not quite possible, at least something that works on *all* subsets of $\RR$. We'll come back to this in a few lectures. ## Notions of "smallness" in $\RR$ **Definition**: Let $E$ be a set, then $E$ is *countable* if it is in a one-to-one correspondence with $E' \subseteq \NN$, which includes $\emptyset, \NN$. **Definition**: A set $E$ is *meager* (or of *1st category*) if it can be written as a countable union of **nowhere dense** sets. *Exercise*: Show that any finite subset of $\RR$ is meager. Intuitively, a set is *nowhere dense* if it is full of holes. Recall that a $X \subseteq Y$ is dense in $Y$ iff the closure of $X$ is all of $Y$. So we'll make the following definition: **Definition**: A set $A \subseteq \RR$ is *nowhere dense* if every interval $I$ contains a subinterval $S \subseteq I$ such that $S \subseteq A^c$. Note that a *finite* union of nowhere dense sets is also nowhere dense, which is why we're giving a name to such a *countably infinite* union. *Example:* $\QQ$ is an infinite, countable union of nowhere dense sets that is not itself nowhere dense. Equivalently, - $A^c$ contains a dense, open set. - The interior of the closure is empty. We'd like to say a set is *measure zero* exactly when it can be covered by intervals whose lengths sum to less than $\varepsilon$ for any $\varepsilon > 0$. **Definition**: $E$ is a *null set* (or has *measure zero*) if $\forall \varepsilon >0$, there exists a sequence of intervals $\theset{I_j}_{j=1}^\infty$ such that $$ E \subseteq \union_{j=1}^\infty \text{ and } \sum \abs{I_j} < \varepsilon. $$(A second proof of A): *Exercise*: Show that a countable union of null sets is null. We have several relationships - Countable $\implies$ meager, but not the converse. - Countable $\implies$ null, but not the converse. *Exercise*: Show that the "middle third" Cantor set is not countable, but is both null and meager. Key point: the Cantor set does not contain any intervals. **Theorem**: Every $E \subseteq \RR$ can be written as $E = A \disjoint B$ where $A$ is null and $B$ is meager. > This gives some information about how nullity and meagerness interact -- in particular, $\RR$ itself is neither meager nor null. > Idea: if meager $\implies$ null, this theorem allows you to write $\RR$ as the union of two null sets. This is bad! *Proof*: We can assume $E = \RR$. Take an enumeration of the rationals, so $\QQ = \theset{q_j}_{j=1}^\infty$. Around each $q_j$, put an interval around it of size $1/2^{j+k}$ where we'll allow $k$ to vary, yielding multiple intervals around $q_j$. To do this, define $$ I_{j, k} = (q_j - 1/2^{j+k}, q_j + 2^{j+k}) .$$ Now let $G_k = \union_j I_{j, k}$. Finally, let $A = \intersect_k G_k$; we claim that $A$ is null. Note that $\sum_j \abs{I_{j, k}} = \frac{1} {2^k}$, so just pick $k$ such that $\frac 1 {2^k} < \varepsilon$. Now we need to show that $A^c \definedas B$ is meager. Note that $G_k$ covers the rationals, and is a countable union of open sets, so it is dense. So $G_k$ is an open and dense set. By one of the equivalent formulations of meagerness, this means that $G_k^c$ is nowhere dense. But then $$ B = \union_k G_k^c $$ is meager. ## $\RR$ is not small **Theorem A**: $\RR$ is not countable. **Theorem B (Baire Category)**: $\RR$ is not meager. **Theorem C**: $\RR$ is not null. Note that theorems B and C imply theorem A. You can also replace $\RR$ with any nonempty interval $I = [a,b]$ where $a< b$, which is a strictly stronger statement -- if any subset of $\RR$ is not countable, then certainly $\RR$ isn't, and so on. *Proof of (A):* Begin by thinking of $I = [0,1]$, then every number here has a unique binary expansion. So we are reduced to showing that the set of all Bernoulli sequences (infinite length strings of 0 or 1) is uncountable. Then you can just apply the usual diagonalization argument by assuming they are countable, constructing the table, and flipping the diagonal bits to produce a sequence differing from every entry. $\qed$ *A second proof of (A)* Take an interval $I$, and suppose it is countable so $I = \theset{x_i}$. Choose $I_1 \subseteq I$ that avoids $x_1$, so $x_1\not\in I_1$. Choose $I_2 \subseteq I_1$ avoiding $x_2$ and so on to produce a nested sequence of closed intervals. Since $\RR$ is complete, the intersection $\intersect_{n=1}^\infty I_n$ is nonempty, so say it contains $x$. But then $x\in I_1 \in I$, for example, but $x\neq x_i$ for any $i$, so $x\not\in I$, a contradiction. $\qed$ *Proof of (B):* Suppose $I = \union_{i=1}^\infty A_n$ where each $A_n$ is nowhere dense. We'll again construct a nested sequence of closed sets. Let $I_1 \subseteq I$ be a subinterval that misses all of $A_1$, so $A_1 \intersect I_1 = \emptyset$ using the fact that $A_1$ is nowhere dense. Repeat the same process, let $I_2 \subset I_1 \setminus A_2$. By the nested interval property, there is some $x\in \intersect A_i$. $\qed$ > Note that we've constructed a meager set here, so this argument shows that the **complement of any meager subset of $\RR$ is nonempty**. > Setting up this argument in the right way in fact shows that this set is dense! Taking the contrapositive yields the usual statement of Baire's Category Theorem. ## Discontinuities Consider the Thomae function: it is continuous on $\QQ$, but discontinuous on $\RR\setminus\QQ$. Can this be switched to get some function $f$ that is continuous on $\RR\setminus \QQ$ and discontinuous on $\QQ$? The answer is no. The set of discontinuities of a function is *always* an $F_\sigma$ set, and $\RR\setminus \QQ$ is not an $F_\sigma$ set. Equivalently, the rationals are not a $G_\delta$ set. Let $D_f$ denote the set of discontinuities of $f$. **Some facts:** - For The pointwise limit of continuous functions, $D_f$ is meager. - If $f$ is integrable, $D_f$ is null. - If $f$ is monotone, $D_f$ is countable. - There is a continuous nowhere differentiable function: - Let $$ f(x) = \sum_n \frac{\norm{10^n x}}{10^n}, $$ and in fact *most* functions are like this. - If $f$ is continuous and monotone, $D_f$ is null. **Theorem**: Let $I = [a,b]$. Then $$ I \subseteq \union_{i=1}^\infty I_i \implies \abs{I} \leq \sum_{i=1}^\infty \abs{I_i} .$$ *Proof*: The proof is by induction. Assume $I \subseteq \union_n^{N+1} I_n$, where wlog we can assume that $a < a_{N+1} < b \leq b_{N+1}$, then $[a, a_{N+1}] \subset \union_{n=1}^N I_n$ so the inductive hypothesis applies. But then $$ b-a \leq b_{N+1} - a = (b_{N+1} - a_{N+1}) + (a_{N+1} - a) \leq \sum_{n=1}^{N+1} \abs{I_n} .$$ $\qed$ > Note that this proves that $\RR$ is uncountable!