13. Comments about Homework 6
≪ 12. Derivatives | Table of Contents | 14. Contraction Mappings ≫I’ve finished grading the sixth homework. I wanted to point out a couple of relatively common mistakes on the second and third problems, which state:
Problem 2. (Discontinuities of monotone functions)
Let \(f:\left[ a,b \right]\to \mathbb{R}\) be monotone nondecreasing. Show that \(f\) has countably many discontinuities.
Problem 3. (Monotone functions are integrable)
Let \(f:\left[ a, b \right]\to \mathbb{R}\) be monotone nondecreasing. Show that \(f\) is Riemann integrable.
For Problem 2, many of you had the right intuitive idea — \(f\)’s points of discontinuity are “upward jumps”, and it cannot jump up by more than \(f(b)-f(a)\) by monotonicity. However, it is important to argue this point precisely, and to do so, one must be able to mathematically quantify the size of the “jump” at a point in terms of the function \(f\). The right thing to do here is to define the “jump” at a point \(x\) as the quantity \[\inf \left\lbrace f(y) : y > x \right\rbrace - \sup \left\lbrace f(y) : y < x \right\rbrace.\] When \(x=a\) or \(x=b\), of course, one needs to replace the \(\sup\) with \(f(a)\) or the \(\inf\) with \(f(b)\), respectively. From here, one may argue as follows:
- The jump at \(x\) is nonzero if and only if \(f\) is discontinuous at \(x\).
- The set of points at which the jump is larger than \(\frac{1}{n}\) is finite for all \(n\in \mathbb{N}\).
- The set of points of discontinuity is the union of the above (finite!) sets as \(n\) ranges over all natural numbers.
For Problem 3, there were several different kinds of errors people made. The most common error, and by far the most critical error, was as follows:
- Partition \([f(a), f(b)]\) into intervals of width no greater than \(\frac{1}{n}\).
- Partition \([a, b]\) by taking preimages of the subintervals from step 1 under \(f\).
- Compute \(U(f, P)-L(f, P)< \frac{b-a}{n}\), where \(P\) is the partition constructed in step 2. Taking \(n\to\infty\) concludes the proof.
The error here is actually very subtle. This construction suggests that the oscillation of \(f\) (the difference between the supremum and the infimum) in each interval of the partition is less than \(\frac{1}{n}\), but this is not the case. Consider the function \(f:\left[ 0,1 \right]\to \mathbb{R}\) given by \[f(x) = \begin{cases} 0 & x < \frac{1}{2} \\ 1 & x \geq \frac{1}{2}.\end{cases}\] When \(n=2\), one partitions the image of \(f\) into \(\left[ 0, \frac{1}{2} \right]\cup \left[ \frac{1}{2},1 \right]\). Taking preimages, one obtains the intervals \(\left[ 0, \frac{1}{2} \right)\cup \left[ \frac{1}{2}, 1 \right]\). So, our partition is \(x_0 = 0, x_1 = \frac{1}{2}, x_2 = 1\).
Remark 4.
Some solutions defined the endpoints of these intervals in a slightly different manner: for instance, one may consider \(y_0=0, y_1=\frac{1}{2}, y_2=1\) as the partition of the image of \(f\). Then, one defines \[x_i = \inf \left\lbrace x : f(x) \geq y_i \right\rbrace.\] However, this now yields \(x_0=0\) and \(x_1=x_2=\frac{1}{2}\), which is not a partition of \([0, 1]\)!
But the problem now becomes: \[\begin{align*} \sup _{x\in \left[ 0, \frac{1}{2} \right]} f(x) = 1 && \textrm{while} && \inf _{x\in \left[ 0,\frac{1}{2} \right]}f(x) = 0. \end{align*}\] Here, the oscillation of \(f\) is no longer bounded by \(\frac{1}{2}\), invalidating the estimate that \(U(f, P)-L(f, P)<\frac{b-a}{2}\). (Although this inequality does end up being true in this case, one can make it false by modifying \(f\) to be \(f(x)=2\) when \(x>\frac{1}{2}\). The point is that the reasoning is flawed!)
The key takeaway is: this approach attempts to construct a partition where the oscillation on every subinterval is arbitrarily small. However, this is almost never possible when there are points of discontinuity.
A more robust tactic was to separate off the points of discontinuity where the “jumps” were large in very thin intervals, then figure out a way to partition the rest of the interval whene the jumps were small. This is a good tactic, and it gives a good picture as to why these functions should be integrable at all. The work can get very messy, and the best way to work around this is by creating a uniform partition and use monotonicity to create a telescoping sum. You can find such a solution on StackExchange.
Finally, I should point out that some arguments were contingent upon the continuity of \(f\) on some sub-interval. These arguments do not work, as there are monotone functions that are discontinuous on a countable, dense subset of their domain!
Exercise 5.
Recall that the rationals are countable; fix an enumeration \(\left\lbrace q_1, q_2, q_3, \ldots \right\rbrace\) of the set \(\mathbb{Q}\cap \left[ 0, 1 \right]\). Define the function \[f(x) = \sum _{n : q_n \leq x} 2 ^{-n}.\]
- Show that \(f\) is well-defined at every point in \([0, 1]\).
- Show that \(f\) is strictly increasing.
- Show that \(f\) is continuous at a point \(x\) if and only if \(x\) is irrational. Conclude that the set of discontinuities of \(f\) is countable and dense in \([0, 1]\).
- Show that \(f\) is Riemann integrable.