Intuition on theorem from Calculus of manifolds Spivak

Question

In his book, Calculus of manifolds Spivak gives following theorem about oscillations:

Let $A \subset \mathbb R^n $ be closed. If $ f:A \rightarrow \mathbb R$ is any bounded function, and $\epsilon >0, $ then $ \{x\in A : o(f,x) \geq \epsilon\}$ is closed.

The quantity $o(f,x)$, known as the oscillation of $f$ at $x$, is defined as $$ \lim_{\delta\to 0}\bigl(M(x,f,\delta)-m(x,f,\delta)\bigr) \, , $$ where $M(x,f,\delta)$ is the supremum of $f(z)$ for $|z-x|<\delta$, and $m(x,f,\delta)$ is the infimum. (It is implicit that $\delta>0$, and so the limit above is a right-hand limit.)

What is the implication of this theorem? What does it mean intuitively? I understand the theorem but want a feel for it.

Answer 1

Let $D_n = \{x \in A: \, o(f,x) \geqslant 1/n\}$.

A function $f$ is continuous at $x$ if and only if the oscillation $o(f,x) = 0$. Hence, the set of points in $A$ where $f$ is discontinuous is

$$D_f = \bigcup_{n \in \mathbb{N}}D_n ,$$

and, with the $D_n$ closed, this leads to the useful characterization of $D_f$ as an $\mathcal{F}_\sigma$ set.

There are many implications of the fact that $D_n$ is closed, including two important theorems.

If a bounded function is almost everywhere continuous, then it is Riemann integrable.

If $f$ is a pointwise limit of a sequence of continuous functions, then it is of Baire class 1 -- meaning the points of continuity of $f$ form an everywhere dense set.

The second theorem is important in that it shows that while a sequence of continuous functions can converge to a discontinuous function, the set of points where it is discontinuous is small in some sense (meager).

Answer 2

I recently came up with the following proof of this theorem that is different to the one given in Spivak. It uses the following characterisation of closed sets: $E$ is closed if and only if every limit point of $E$ is a point in $E$.

Suppose $p\in\mathbb R^n$ is a limit point of $B=\{x\in A:o(f,x)\ge\varepsilon\}$. We claim that $p\in B$. To begin with, $p\in A$, since $p$ is a limit point of $A$ and $A$ is closed. Now, for every $\delta>0$ we may choose an $x_{\delta}\in B$ such that $0<|p-x_{\delta}|<\delta/2$. If $\delta>0$, then it follows from the triangle inequality that the open ball centred at $x_{\delta}$ with radius $\delta/2$ is a subset of the open ball centred at $p$ and with radius $\delta$. Hence, $$ M(p,f,\delta)-m(p,f,\delta)\ge M(x_{\delta},f,\delta/2)-m(x_{\delta},f,\delta/2)\ge\varepsilon \, . $$ Since $M(p,f,\delta)-m(p,f,\delta)$ is always $\ge\varepsilon$, its limit as $\delta\to0$ must also be so. This completes the proof.

For me, the intuition behind this proof is that because we can find points arbitrarily close to $p$ with oscillation $\ge\varepsilon$, we can use those points to estimate the oscillation of $f$ at $p$. That being said, I don't find this theorem particularly intuitive – it took me a lot of false starts to come up with the proof above.