Home Math Teaching About Me

7. Week 6: Integrals and Meshes

≪ 6. Week 5: Integrals and Limits | Table of Contents | 8. Week 7: Metrising the Space of Continuous Functions ≫

I suppose I should allot some time for midterm review things…

One problem from the practise midterm I’d like to highlight is the following:

In other words, there is no continuous bijection $f: [0, 1]\to \left[ 0,1 \right]^2$. A highlight of this problem is the following fact: if $f$ is continuous and bijective on a compact domain, its inverse is also continuous. This is not true of non-compact domains — make sure you know some counterexamples.

Since continuous functions preserve topological features in one direction or another, continuous bijections with continuous inverses (known as homeomorphsims) identify topological features of the domain and range: open sets and closed sets are “paired up”, connectedness for one implies connectedness of the other, etc. With this in mind, one may search for topological features of $[0, 1]$ that are not shared by $[0, 1]^2$ to get a contradiction.

Solution 1

If $f : [0, 1]\to \left[ 0,1 \right]^2$ is a continuous bijection, its inverse $f ^{-1} : \left[ 0,1 \right]^2 \to \left[ 0,1 \right]$ is continuous too. Restricting this to the smaller domain $$f ^{-1} : \left[ 0,1 \right]^2\setminus \left\lbrace f ^{-1}\left( \frac{1}{2} \right) \right\rbrace \to \left[ 0,1 \right]\setminus \left\lbrace \frac{1}{2} \right\rbrace$$ gives a continuous surjection from a connected domain to a disconnected image, a contradiction. $\square$

This solution argues that the unit square and unit interval are topologically distinguishable, as the latter becomes disconnected upon removing a point while the former remains connected.

Here’s another cute solution that shows that the unit square and unit cube are topologically distinguishable by size:

Solution 2

Suppose $f : [0, 1]\to \left[ 0,1 \right]^2$ is a continuous bijection; its inverse is also continuous. Then, if $\iota _x : (0, 1)\to [ 0, 1]^2$ is given by $y\mapsto (x, y)$, we have the composite $$(0, 1) \xrightarrow{\iota _x} \left[ 0,1 \right]^2 \xrightarrow{f ^{-1} } [0, 1]$$ is continuous and injective. Hence, by the intermediate value theorem, $f ^{-1} \left( \left\lbrace x \right\rbrace\times (0, 1) \right)$ is a nonempty open interval, and these are pairwise disjoint. But $[0, 1]$ does not contain uncountably many nonempty disjoint open intervals. $\square$

The Mesh Size Characterisation of the Integral

Let $\alpha : [a, b]\to \mathbb{R}$ be a nondecreasing function. Recall the following criterion for Riemann-Stieltjes integrability with respect to $\alpha$:

Let $f : [a, b]\to \mathbb{R}$ be continuous. Recall the following proof that $f \in \mathscr{R}(\alpha )$:

1. Since $f$ is continuous on a compact domain, it’s uniformly continuous. Thus, given $\epsilon > 0$, let $\delta > 0$ such that $\left\lvert x-y \right\rvert < \delta$ implies $\left\lvert f(x) - f(y) \right\rvert < \frac{\epsilon }{ \alpha (b) - \alpha (a)}$.
2. Let $\mathcal{P} = \left\lbrace a = x_0 < \cdots < x_n =b \right\rbrace$ be any partition of $[a, b]$ such that $x_i - x _{i-1} < \delta$ for all $i$. On each subinterval $\left[ x _{i-1}, x_i \right]$, we have $M_i - m_i < \frac{\epsilon }{\alpha (b) - \alpha (a)}$ by the extreme value theorem and by construction of $\delta$.
3. Compute that $$U(f, \mathcal{P}, \alpha ) - L (f, \mathcal{P}, \alpha ) = \sum _{i=1}^{n} \left( M_i - m_i \right) \Delta \alpha _i < \frac{\epsilon }{ \alpha (b) - \alpha (a)} \sum _{i=1}^{n} \Delta \alpha _i = \epsilon .$$

Note that this proof indicates a bit more than integrability: it doesn’t matter what the partition $\mathcal{P}$ is, as long as it’s “fine enough”, the upper and lower sums will be good approximations of the integral. This is in line with how we learned integration in calculus. This can be formalised as follows:

Explicitly, this means that for all $\epsilon > 0$, there exists a $\delta > 0$ such that $\mu (\mathcal{P}) < \delta$ implies $$\left\lvert \int _{a}^{b} f\ d \alpha - U \left( f, \mathcal{P}, \alpha \right)\right\rvert < \epsilon ,$$ and likewise for the lower sums.

This leads to the following alternative definition of integrability: if $f: [a, b]\to \mathbb{R}$ is bounded, $f\in \mathscr{R} (\alpha )$ if the limits $$\lim _{\mu (\mathcal{P}) \to 0} U \left( f, \mathcal{P}, \alpha \right) = \lim _{\mu \left( \mathcal{P}\right)\to 0 } L \left( f, \mathcal{P}, \alpha \right)$$ and both exist. Unfortunately, this is not equivalent to the definition of integrability we stated at the beginning.

As your condition hopefully demonstrates, it’s when the discontinuities of $\alpha$ and $f$ overlap and “conspire together” that the mesh size formulation of integrability fails. When $f$ is continuous, this pathology doesn’t happen. When $\alpha$ is continuous, we again expect these pathologies not to happen either, but proving the equivalence of these two characterisations is more difficult.

Note, by the way, that the limits as $\mu (\mathcal{P}) \to 0$ not existing or not coinciding does not imply that both $f$ and $\alpha$ share a point of discontinuity. Any $f\notin \mathscr{R}(\alpha )$ exhibit this problem, so this strategy of proof is doomed.

To prove the proposition, we’ll first prove a lemma that states that upper and lower sums are “continuous” with respect to $\mathcal{P}$ when $\alpha$ is continuous. Specifically, when you add a point to a partition, the upper and lower sums do not change by much.

To prove this key lemma, expand the two sums and compare.

Proof of Proposition

The “if” direction is clear: take any partition of sufficiently small mesh size, and the upper and lower sums will be close to their common limits, hence also each other.

We will focus on the “only if” part and prove that if $f\in \mathscr{R}(\alpha )$, then $$\lim _{\mu (\mathcal{P})\to 0} U\left( f, \mathcal{P}, \alpha \right) = \int _{a}^{b} f\ d \alpha.$$ An identical argument applies to the limit of the lower sums.

When $f$ is a constant function, both claims are trivially true, so we’ll assume $f$ is not constant.

Fix any $\epsilon _0 > 0$. There exists a partition $\mathcal{P}$ such that $U \left( f, \mathcal{P}, \alpha \right) < \int _{a}^{b} f\ d \alpha + \epsilon _0$. Our goal is to determine a $\delta > 0$ such that whenever $\mu \left( \mathcal{P}’ \right) < \delta$, $U \left( f, \mathcal{P}’, \alpha \right)$ is within $\epsilon _0$ of the integral (maybe with an extra constant factor).

To do so, we’ll leverage the lemma and argue that $U \left( f,\mathcal{P}’, \alpha \right)$ should be close to $U \left( f, \mathcal{P}’\cup \mathcal{P}, \alpha \right) \leq U \left( f, \mathcal{P}, \alpha \right)$. We can do this iteratively, adding one point at a time to $\mathcal{P}’$, leveraging the continuity of $\alpha$ to make the difference at each step very small.

Let $\epsilon _1 > 0$, to be chosen later. Since $\alpha$ is continuous on a compact domain, it’s uniformly continuous. Thus, there exists $\delta > 0$ such that $\left\lvert x-y \right\rvert < \delta$ implies $\left\lvert \alpha (x) - \alpha (y) \right\rvert < \epsilon _1$. Then, whenever $\mathcal{P}’$ is any partition with $\mu \left( \mathcal{P}’ \right) < \delta$, one has $$\begin{align*} U\left( f, \mathcal{P}’, \alpha \right) - U \left( f, \mathcal{P}’ \cup \left\lbrace t_* \right\rbrace , \alpha \right) &\leq \left( \sup f - \inf f \right) \cdot \left( \alpha \left( x_i \right)- \alpha \left( x _{i-1} \right)\right) \\ &< \left( \sup f - \inf f \right) \epsilon _1 . \end{align*}$$ Proceeding by induction, we have that $$U \left( f, \mathcal{P}’, \alpha \right) - U \left( f, \mathcal{P}’ \cup \mathcal{P}, \alpha \right) < \left( \sup f - \inf f \right) \epsilon _1 \left\lvert \mathcal{P} \right\rvert,$$ where $\left\lvert \mathcal{P} \right\rvert$ is the number of points in the partition $\mathcal{P}$.

Setting $\epsilon _1 = \frac{\epsilon _0}{ \left\lvert \mathcal{P} \right\rvert \left( \sup f - \inf f \right)}$, we find that $$\begin{align*} U \left( f, \mathcal{P}’, \alpha \right) &< U \left( f, \mathcal{P}’\cup \mathcal{P} , \alpha \right) + \epsilon_0 \\ &\leq U \left( f, \mathcal{P}, \alpha \right) + \epsilon _0 \\ &< \int _{a}^{b} f\ d \alpha + 2\epsilon _0. \end{align*}$$ This concludes the proof. $\square$

Think carefully about the order in which everything was constructed and chosen:

1. $\epsilon_0 > 0$ was taken to be arbitrary.
2. $\mathcal{P}$ was determined based on $\epsilon _0$.
3. $\epsilon _1 > 0$ was determined based on $\mathcal{P}$, $f$, and $\epsilon _0$.
4. Finally, $\delta > 0$ was determined based on $\epsilon _1$ and $\alpha$.

The lemma is suggestive of why we need either $f$ or $\alpha$ to be continuous in order to make the above statement: when we relate fine partitions to “coarse” partitions, we need to make sure the two are close. When both $f$ and $\alpha$ are simultaneously discontinuous, it’s possible that this doesn’t happen.