Math3033 (2013-2014) Final Exam

This year the mean of the final exam is 35.51/100 and the SD is 12.71. To wit, answering the first two problems perfectly is already safe to get $\text{C+}/\text{B-}$. If one does perfectly well in everything before the final exam (which are all very standard), then one gets an $\text{A-}$. This again reflects that there are too many weak students in Math3033 and getting an $\text{A-}$ is very easy for hard working students.

The first two problems are routine calculation, with one problem on uniform convergence and one on Lebesgue Dominated Convergence Theorem. The next 3 problems are as follows:

Problem 3. Show that the function $f:[0,1]\setminus \Q\to \R$ given by \[
f(x)=\sum_{i=1}^\infty \frac{a_i}{16^i},\quad\text{where } x=0.a_1a_2a_3\dots
\] is measurable.

Summary to Problem 3. Very few people can get it correct, some partial marks (up to 2) are given for those who try to prove $f$ is continuous, increasing, etc. Indeed this function is not continuous, any attempt proving this to be continuous just deserve very little marks. Some of the students claim the function is increasing, this still requires some manipulations of inequality and cannot be seen as trivial.$\qed$

Problem 4. Let $f:[0,1]\to[0,\infty)$ be measurable such that \[
\int_{[0,1]} f^n\, dm =c\in \R,\quad \forall n\in \N.
\] (a) Prove that $f^{-1}[2013,\infty)$ has measure zero.
(b) Prove that $f=\chi_E$ a.e. for some measurable set $E$.

Summary to Problem 4. By observing the answer in part (b), some students can guess the answer in part (a) but did not attempt to prove this. Actually it is very easy to show $m(f^{-1}(\alpha,\infty))=0$ for every $\alpha > 1$, therefore $f\leq 1$ a.e. The part (b) is hard to many students, they need to explain why $m(f^{-1}[0,1))=0$, this should not be difficult if one tries to play around with the term $\int_{[0,1]}f^n\,dm = c$ and split the integral into two parts. Then \[
m(f^{-1}(\{1\}))=1,
\] and therefore on $f^{-1}(\{1\})$ we have $f = 1=\chi_{f^{-1}(\{1\})}$. It follows that $f=\chi_E$ a.e., where $E=f^{-1}(\{1\})$.$\qed$

Problem 5. Let $T\subseteq [0,1]$ be a set of $7$ elements and $S\subseteq [0,1]$ be measurable such that $m(S)\ge \frac{1}{3}$, prove that there are $a,b\in T$ and $c,d\in S$ such that \[
|a-c|=|b-d|.
\] 

Summary to Problem 5. Only 2 people can get a decent amount of marks from me. Both of them try to prove by contradiction by supposing that $|a-c|\neq |b-d|$ for every $a,b\in T$ and $c,d\in S$. One of them can proceed successfully while another one is on the half way. Students usually work on equivalent statement of $|a-c|=|b-d|$ but never think of the possibility of having the sufficient condition: \[
a-c=b-d
\] for some $a,b\in T$ and $c,d\in S$. In this way one can find that \[
a+d=b+c.
\] Namely, we may try to prove $(a+S)\cap( b+S)\neq \emptyset$ for some $a,b\in T$. Suppose not, the we have the contradiction that $2\ge 7/3$.$\qed$

Conclusion. It seems that Dr Li thinks the midterm is easy enough to ensure most people can pass the course, therefore it tries to set the exam in a moderate level. Unfortunately very few students can develop a good background (absorb the material) and technique (methodology) in the basic measure theory.