Add homework/exam feedback

analysis--berkeley-202a-final.tex

@@ -3,6 +3,8 @@ \section*{Math 202A - Final Exam - Dan Davison - \texttt{[email protected]}}
 \begin{mdframed}
   \includegraphics[width=400pt]{img/analysis--berkeley-202a-final-c9d2.png}
 \end{mdframed}
+
+
 \green{COMPLETE}
 
 \begin{proof}
@@ -436,6 +438,54 @@ \section*{Math 202A - Final Exam - Dan Davison - \texttt{[email protected]}}
 \begin{mdframed}
   \includegraphics[width=400pt]{img/analysis--berkeley-202a-final-21a6.png}
 \end{mdframed}
+% Problem 4
+% How did people do 4 on final? I don't know how to say anything about the "singular continuous case", e.g. if μ is Cantor-Lebesgue measure.
+
+% Or in other words what I mean is, the condition is pretty obvious if we are testing at a point x such that μ({x})>0, because the numerator of the limit is bounded below when ε→0. We only need to check μ-almost every x, so this sort of close to solving it but not there, because there could be an uncountable, Lebesgue measure zero set T such that μ(x)=0 for all x∈T but μ(T)>0. Indeed this is the case if μ is the measure derived from the Cantor-Lebesgue function; I think T could be any uncountable subset of the Cantor set, and certainly the whole thing. What can we say about that quotient if x∈T?
+
+
+% exam
+% edit·good question1Updated 8 hours ago by HUGO A JENKINS
+% the students' answer,where students collectively construct a single answer
+% Click to start off the wiki answer
+% followup discussionsfor lingering questions and comments
+% Resolved Unresolved
+% Anonymous Scale
+% Anonymous Scale 8 hours ago
+% I was only able to show that limsupϵ>0(μ([x−ϵ,x+ϵ])2ϵ)=+∞ for μ-almost every x∈R, so I would also be curious to see how others did it.
+
+% What I did was take an arbitrary α>0, then for each n∈N, cover the Lebesgue measure zero set whose complement is μ measure zero with intervals of total length less than 4−nα, throw out all intervals with μ value less than their length multiplied by 2n, and let An be the union of the remaining intervals. Then, let A be the intersection of all An. At this point, μ(AC)<α, and for any x∈A, the limsup goes to infinity. Then because α is arbitrary, the limsup goes to infinity μ-almost everywhere. I couldn't find a way to control the liminf, however.
+
+% helpful! 0
+% Reply to this followup discussion
+% Resolved Unresolved
+% HUGO A JENKINS
+% HUGO A JENKINS 8 hours ago
+% How do you have μ(Ac)<α? Or even μ(Acn)<α? I do not see how α controls μ of anything; as in when you throw away intervals, you are only adding μ-value to these complements
+
+% helpful! 0
+% Anonymous Scale
+% Anonymous Scale 7 hours ago The total length of the intervals is less than 4−nα, so when you throw out the intervals whose μ value is less than 2n times their length, the total μ value you've gotten rid of is less than 2−nα.
+% helpful! 1
+% HUGO A JENKINS
+% HUGO A JENKINS 7 hours ago
+% Ah I see. Great.
+
+% But why does the limsup go to infinity for any x∈A? Could x just happen to be in every An, but if you center at x and then take the limit it stays bounded….?
+
+% helpful! 0
+% HUGO A JENKINS
+% HUGO A JENKINS 7 hours ago
+% Ah no I see, sorry. Great.
+
+% Maybe we can institute more strict requirements to not be chucked out when forming An. Like neither of your left or right half subintervals can have μ-density less than 2n
+% helpful! 0
+% Harsh Srivastav
+% Harsh Srivastav 3 hours ago I used a similar treatment as Bass Chapter 14.3, essentially replacing the lim sup with a lim inf and replacing λ with m wherever I found it, as well as the appropriate conditions; it seemed to give me what I wanted.
+% helpful! 0
+
+
+
 \red{INCOMPLETE}
 
 
@@ -737,6 +787,18 @@ \section*{Math 202A - Final Exam - Dan Davison - \texttt{[email protected]}}
 \begin{mdframed}
   \includegraphics[width=400pt]{img/analysis--berkeley-202a-final-8aed.png}
 \end{mdframed}
+
+% You want to make use of the fact that the complement of $S$ is open in $\mathbb R$, meaning
+% that it is a disjoint finite or countable union of open intervals, say $(a_n, b_n)$. Let $P_n$
+% be the preimage under $f$ of $(a_n, b_n)$. Then we must have $\mu(P_n) = 0$, as otherwise,
+% the given condition (only applicable to sets with $\mu > 0$) would give us
+% $a_n < \frac1{\mu(P_n)} \int_{P_n} f \, d\mu < b_n$.
+
+
+
+
+which contradicts that $$A_{P_n}(f) \in S$$. Since the preimage of $$(a_n, b_n)$$ under $$f$$ has measure 0, the preimage of the union of $$(a_n, b_n)$$ over $$n$$ has measure 0 as well, whence $$f(x) \in S$$ for $$\mu$$-almost all $$x$$.
+
 \red{INCOMPLETE}
 
 First, we prove this for $X = \R^n$.

analysis--berkeley-202a-hw12.tex

@@ -1,5 +1,20 @@
 \section{Math 202A - HW12 - Dan Davison - \texttt{[email protected]}}
 
+% 3. Your intuition is correct and suggests a proof a little different from the one that other
+% students have been given: by an approximation argument, I think we can assume that f has compact
+% support, then just take x to be far outside the support of f and compute the maximal function at x.
+% (-6) Aidan Backus, Dec 2 at 1:47pm
+
+% 2. You're right that this doesn't quite work when n > 1 (for one, your definition of V isn't quite
+% corect) but you're right to consider that B(x, 2r) \supseteq B(x', r) where |x' - x| < r. (-4)
+
+% 3. Yes, you did. Otherwise you don't know that the integral exists for sufficiently small r.
+% Ian Francis, Dec 20 at 10:58pm
+
+% 1. 4 2. 10 3. 6 5. 0
+% Ian Francis, Dec 21 at 12:23am
+
+
 % Recall that for locally integrable $f$, the average value of $f$ on a ball centered at $x$ is
 % \begin{align*}
 %   (A_r f)(x) = \frac{1}{m(B(x, r))} \int_{B(x, r)} f,

analysis--berkeley-202a-hw13.tex

@@ -235,8 +235,37 @@ \section{Math 202A - HW13 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw13-9356.png}
 \end{mdframed}
 
-First, let's examine some small finite sets and the possible topologies that meet the specified condition. The
-following table excludes topologies that differ only by a relabeling of the elements in the underlying set.
+% From Aidan Backus (grader):
+
+% Recall that a directed graph is "transitive" if given edges x→y→z we can compose them to get an
+% edge x→z. The key point is that if X is a finite topological space then we can think of X as (the
+% vertices of) a transitive directed graph. Namely, if x,y∈X then we define an edge x→y iff x is in
+% the closure of y. Conversely, if X is a transitive directed graph we can put a topology on (the
+% vertices of) X by saying that the open sets are the sets which are closed upwards.
+
+% The following are equivalent: X is connected (in the sense of topological spaces); and X is
+% graph-theoretically connected. Indeed, every graph-theoretically connected component of X is a
+% closed subset of X, so it is clopen since its complement is closed; the converse is similar.
+
+% If X is as in the problem and there is an edge x→y then either x=y or there is no edge y→x. Suppose
+% that there are edges x→y→x. Then x, y have the same closure, so x, y are indiscernible. Therefore
+% {x} is not open, so it is closed, so it is the closure of x, so x=y.
+
+% Suppose that every proper nonempty subset of X is either open or closed but not open. Then X has no
+% proper nonempty clopen subsets, so X is connected, and hence graph-theoretically connected. By the
+% previous paragraph, X has no cycles except for the trivial edges x→x. I think that one can now show
+% that there is a unique vertex x such that either x is initial or x is final. To do this, you first
+% might show that X is the transitive closure of a directed tree.
+
+% If x is initial, then x is in the closure of every point. So every set that contains x is open and
+% every other set is closed. Otherwise, x is final; by symmetry (since we're in a finite topological
+% space, the set of closed sets is also a topology!) every set that contains x is closed and every
+% other set is open.
+
+
+First, let's examine some small finite sets and the possible topologies that meet the specified
+condition. The following table excludes topologies that differ only by a relabeling of the elements
+in the underlying set.
 
 [Incomplete, table is incomplete and I didn't figure out what the pattern was.]
 
@@ -334,6 +363,11 @@ \section{Math 202A - HW13 - Dan Davison - \texttt{[email protected]}}
     many circles then set $K_\gamma = \infty$. We have $K_\gamma \geq 1$ because each loop starts
     at $(0, 0)$.
 
+    Note that a loop can follow infinitely many circles without needing to ``move at infinite
+    speed'': for example, viewing $t \in [0, 1)$ as a time parameter, we may divide $t$ up
+    as $\sum_{n=1}^\infty \frac{1}{2^n}$, i.e. specify that the particle spends a fraction $2^{-n}$
+    of the total time traversing the $n$-th circle.
+
     Note that a circle may be followed in one of two ``orientations​'': clockwise or anticlockwise.
     Let $\rho_{\gamma, k}$ be the orientation of the $k$-th circle in loop $\gamma$.