Jump Discontinuity

 A real function $f : (a, b) \rightarrow \mathbb{R}$ is said to be continuous at a point $x_0$ if $\lim_{x \rightarrow x_0} f(x) = f(x_0)$. If $f$ is not continuous at $x_0$, then $x_0$ is said to be a point of discontinuity.  For example consider

$$f_1(x) = \begin{cases}
     1  & \text{if  $x = 0$}\\
    0 & \text{otherwise.}
\end{cases}
$$

$f_1$ has a point of discontinuity at $0$ because $lim_{x \rightarrow 0} f_1(x) = 0 \neq f_1(0) = 1$. Continuity can also fail if the limit does not exists. Consider

$$f_2(x) = \begin{cases}
     1  & \text{if  $x > 0$}\\
    0 & \text{if $x \leq 0$}
\end{cases}
$$

which has a point of discontinuity at $0$. Observe that the sequence $a_n = \frac{(-1)^n}{n} \rightarrow 0$ however, $f_2(a_n) = (-1)^n$ has no limit. Even when no limit exists, we can none the less consider the Left and Right Limits
$\lim_{x \rightarrow x_0 -} f(x) = f(x_0-)$ and $\lim_{x \rightarrow x_0 +} f(x) = f(x_0+)$ respectively. 

If the left and right limits exists at $x_0$ but $f(x_0-) \neq f(x_0+)$ (as in $f_2$) or $f(x_0) \neq f(x_0-), \ f(x+)$ (as in $f_1$),  then $x_0$ is a point of jump discontinuity or jump point. Not all discontinuities are jump discontinuities. For example consider the function $g : \mathbb{R} \rightarrow \mathbb{R}$ where $g(0) = 0$ and $g(x) = sin(1/x)$.

(ex*) Show that neither left nor right limits exist for $g$ at the point $0$.

Before stating the problem, here are the details for how limits are defined. There are two definitions of limits on functions function, one involving intervals on the domain and range and another involving sequences.

The first definition states $lim_{x \rightarrow x_0} f(x) = y_0$ if and only if for each $\epsilon > 0$ there exist $\delta > 0$ such that $|f(x) - y_0| < \epsilon$ for any $x \in (x_0 - \delta, x_0 + \delta)$. Informally a function is continuous if and only if it stays more or less constant when we restrict ourselves to a sufficiently small interval of the domain.

Rather than deal with multiple logical quantifiers, it is sometimes simpler to use the "convergent sequence" definition of a limit. According to this definition, $lim_{x \rightarrow x_0} f(x) = y_0$ if and only if for any sequence $x_n \rightarrow x_0$, $f(x_n) \rightarrow x_0$. This definition is simpler and is especially useful for providing counterexamples, as was done to show $f_2$ is discontinuous, although it is, perhaps, a little less intuitive.

(ex) Show the two definitions equivalent.

Left or Right limits can be defined following either definition, in the first case by simply restricting our delta interval to $(x_0 - \delta, x_0)$ or $(x_0, x_0 + \delta)$, and in the second by requiring $x_n < x_0$ or $x_n > x_0$.

(ex) Show that the two definitions for left limit are equivalent and that if $f(x+) = f(x-)$ then $lim_{x \rightarrow x_0 }f(x)= f(x+) = f(x-)$.

Problem:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Show that $f$ has at most countably many points of jump discontinuity.  

solution

First Post

As part of an effort to continue to study of mathematics now that I am no longer taking classes I have decided to start blogging the mathematics I am studying. I intend for this blog both to provide a resource to others studying mathematics and to serve as a means for receiving feedback on my work. Any corrections, comments, or alternative proofs are welcome.

The general format of a post will a consist of brief introduction to a problem, including a discussion of some of the relevant definitions and theorems, the statement of the problem itself, and separate hint and solutions files.  The problems will mainly be based off standard problems from various math texts, but the introductions and solutions will be my own work unless stated otherwise.  My plan is to write-up two or three problems a week, but I am not going to commit to any deadlines just yet.

Any how without further ado, here is my first problem:

A set $X \subset \mathbb{R}$ is said to be of (Lebesgue) measure zero if for every $\epsilon >0$ there exists a countable set of open intervals $C = \{(a_i, b_i)\}_{i \in \mathbb{N}}$ such that $C$ covers $X$ (i.e $X \subseteq \bigcup_i (a_i, b_i)$) and $\sum_{i=1}^\infty
(b_i -a_i) < \epsilon$. The distance $b_i - a_i$ is referred to as the length of the interval $(a_i, b_i)$.

One can see right away that any finite set of points is measure zero. It isn't much more difficult to see that that the integers are measure zero, as is any countable set of points in $\mathbb{R}$. Even some uncountable sets, such as the Cantor set are of measure zero. This corresponds with our intuition about these sets not taking up space.

Exercise: Show that the interval $[0,1]$ is not measure zero.

Solution File

Questions or Comments are welcome.