TITLE: Is the space of uniformly left continuous functions on [0,1] complete?
QUESTION [9 upvotes]: We'll say that a function on $[0,1]$ is uniformly left continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that $x \in (y - \delta, y)$ implies $|f(x) - f(y)| < \epsilon$ for every $x, y \in [0,1]$.  I want to know if the space of all such functions is complete with respect to the uniform norm.
I'm interested in this space because I suspect that it is the completion of the space of piecewise constant functions which are continuous from the left.
Thanks in advance for the help!
EDIT:
As was pointed out in the comments, what I wrote above is equivalent to uniform continuity.  I think the following revised definition captures what I want.  $f$ is uniformly left continuous if for every $\epsilon$ there is a partition $0 = t_0 < t_1 < \ldots < t_n = 1$ such that $|f(x) - f(y)| < \epsilon$ whenever $t_0 \leq x \leq y \leq t_1$ or $t_i < x \leq y \leq t_{i+1}$ for $i > 0$.  So with this definition the characteristic function of $(1/2,1]$ is uniformly left continuous while the characteristic function of $[1/2,1]$ is not.
Here are the examples that I'm really trying to kill.  Take $f$ to be the function which is 0 on the interval $[0,1/2]$ and let $f(x) = 1/(x-1/2)$ on $(1/2,1]$.  This is left continuous, but it shouldn't be uniformly left continuous.  Even if you insist that $f$ be bounded, you could set $f(x) = \sin(1/(x-1/2))$ for $x$ in $(1/2,1]$ and get a left continuous function which is not uniformly left continuous.  If you can think of a better definition that captures this intuition, let me know.

REPLY [4 votes]: The standard $3\epsilon$ trick works.  Suppose $f_n$ is a Cauchy (in uniform norm) sequence of uniformly left continuous functions on $[0,1]$, and $f$ its pointwise limit.  Given $\epsilon > 0$, take $n$ large enough that
$|f_n(x) - f(x)| < \epsilon/3$ everywhere.  Let $0 = t_0 < \ldots < t_k = 1$ be a partition that satisfies your (revised) definition for the function $f_n$ with $\epsilon$ replaced by $\epsilon/3$.  Then for $t_0 \le x \le y \le t_1$ or $t_i < x \le y \le t_{i+1}$, 
$|f(x) - f(y)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| < \epsilon$.