Sets of Discontinuity

In Chapter 4, we spent much of our time trying to understand and prove continuity of functions. However, as we saw in Section 4.1, discontinuous functions can be even more interesting. Recall the Dirchlet Function, which we saw was discontinuous on all of \mathbb{R}, or well as the modified Dirchlet, which was discontinuous on \mathbb{R}\ \{0\} , and don’t forget the Thomae Function, which was discontinuous on all of \mathbb{Q}. What is going on with these discontinuities, and what does this tell us about the discontinuities of an arbitrary function?

To start, let’s define, for a function f: \mathbb{R} \to \mathbb{R}, the set D_f \subseteq \mathbb{R} as the set of all points where f fails to be continuous. It turns out that this set has a number of really interesting properties for a generic function on the real numbers. In general, we can divide discontinuities of a function at a point c into three categories:

  • Removable discontinuity: when \lim{x \to c} f(x) exists but is a different value from f(c) Example: g(x) = \begin{cases} x & x \neq 1 \\ 0 & x = 1 \end{cases} has a removable discontinuity at x = 1
  • Jump discontinuity: when \lim{x \to c^+} f(x) \neq \lim{x \to c^-} f(x). Example: g(x) = \begin{cases} x & x\leq 1 \\ x+6 & x > 1 \end{cases} has a jump discontinuity at x = 1
  • Essential discontinuity: when \lim{x \to c} f(x) doesn’t exist for some other reason–so at least one or both of the right hand limits either doesn’t exist or is infinite. Example: the function g(x) = \sin{\frac{1}{x}} has an essential discontinuity at the point x = 0.

Now, let’s look at the class of monotone functions. Similarly to how we defined a monotone sequence, we can define a monotone function f: A \to \mathbb{R} as a function that is either increasing or decreasing on A, so either f(x) \leq f(y) (increasing) or f(x) \geq f(y) (decreasing) for x < y \in A. Monotone functions can only have a jump discontinuity, because a removable or essential discontinuity would make the function not monotone on A. With a jump discontinuity, a function can still maintain the property that f(x) \leq f(y) or f(x) \geq f(y) for x < y. From there, we can create a bijection between the set of jump discontinuities of f and a subset of \mathbb{Q} to show that D_f for monotone functions is either countable or finite.

Now that we’ve looked at monotone functions, let’s think about the discontinuity set of a non-monotone function. It turns out that, for any arbitrary function f: \mathbb{R} \to \mathbb{R}, D_f is classified as a F_\sigma set, or a set that can be written as the countable union of closed sets (this proof is outlined on p. 143 of the Abbot textbook). We ran into F_{\sigma} sets in section 3.5 of the Abbot text, where they help us define \mathbb{R} as a complete metric space.

Now you may be wondering how we can determine if a set is a F_{\sigma} set and, since every set of discontinuities for an arbitrary function is a countable union of closed sets, whether we can construct a function that is discontinuous on any subset of \mathbb{R}. At first it might look like it–the Dirchlet function has D_f = \mathbb{R} and the Thomae function has D_f = \mathbb{Q}. Does that mean every subset of \mathbb{R} is a F_{\sigma} set? It turns out not.

Let’s take a look at the irrational numbers. We already know that \mathbb{I} = \mathbb{Q}^c. According to Baire’s Theorem, which is described in more detail in Chapter 3 of the Abbot textbook, if \mathbb{R} is written as a F_{\sigma} set, then at least one of those sets must contain a non-empty open interval. Since \mathbb{Q} can be expressed as the countable union of closed intervals (containing singleton points), it cannot contain any open intervals around those points. If \mathbb{I} were a F_{\sigma} set, then \mathbb{I} could not contain any open intervals, otherwise those intervals would intersect \mathbb{Q} because \mathbb{Q} is dense in \mathbb{R}. Since \mathbb{Q} \cup \mathbb{I} = \mathbb{R}, then \mathbb{R} would contain no open intervals, which contradicts Baire’s Theorem, so \mathbb{I} is not an F_{\sigma} set.

We have shown there is no possible function that could be discontinuous on all of \mathbb{I} but continuous on \mathbb{Q}. This has big implications–given an artibtrary subset of the real numbers, it is not always possible to construct a function that will be discontinuous on that set but continuous on the rest of \mathbb{R}. However, if we are given an arbitrary F_{\sigma} set, we will always able to construct a function with D_f equal to that F_{\sigma} set.

 

Sources: http://math.mit.edu/~jspeck/18.01_Fall%202014/Supplementary%20notes/01c.pdf

Stephen Abbot, Understanding Analysis 2nd Edition

Posted on by Fiona Lynch
Categories: Fall 2016

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