Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b\in\mathbb R$ and $L^\infty([a,b])$ be a space of all bounded functions $f:[a,b]\to\mathbb R$. It is a metric space with a metric function given by $$ d(f,g) = \sup\limits_{x\in[a,b]}|f(x) - g(x)|. $$ Let us say that the function $f\in L^\infty([a,b])$ is in the class $S$, i.e. $f\in S$ if there is a finite partition $$ \mathcal T = \{a = t_0<t_1<\dots<t_n = b\} $$ and a sequence of reals $c_0,\dots,c_{n-1}$ such that $$ f(t) = \sum\limits_{i=0}^{n-2}c_i1_{[t_i,t_{i+1})}(t)+c_{n-1}1_{[t_{n-1},t_n](t)} $$ for all $t\in [a,b]$. I wonder if there is a nice characterization of $\bar S$, the closure of $S$ in $L^\infty([a,b])$. E.g. any continuous on $[a,b]$ function is in $\bar S$ - but it surely is much larger.

share|cite|improve this question
Is the last term $c_{n-1}\mathbf 1_{[t_{n+1},t_n]}(t)$? You could be interested in regulated functions. – Davide Giraudo Feb 26 '12 at 20:37
@DavideGiraudo: thanks for mentioning the typo (I guess you meant $t_{n-1}$, not $t_{n+1}$). Regulated functions form exactly the class I was looking for. Would you put this comment as an answer? – Ilya Feb 26 '12 at 21:55
These are usually called step functions. Simple functions allow characteristic functions of sets that are not intervals. – Jonas Meyer Feb 27 '12 at 5:04
@Jonas: thank you! – Ilya Feb 27 '12 at 7:59
up vote 4 down vote accepted

The closure for the uniform norm of step functions is the space of functions which have left and right limit at each point of $(a,b)$ (right for $a$ and left for $b$). These functions are called regulated functions.

If $f$ is in the closure for the uniform norm of step functions, we fix $\varepsilon>0$. We can find a step function $f_1$ such that $||f-f_1||_{\infty}\leq \varepsilon$. Let $x_0\in (a,b)$. Then we can find $\eta>0$ such that $f_1$ is constant on $(x_0,x_0+\eta)$. If $x,y\in (x_0,x_0+\eta)$ then $$|f(x)-f(y)|\leq |f(x)-f_1(x)|+|f_1(x)-f_1(y)|+|f_1(y)-f(y)|\leq 2\varepsilon,$$ so by Cauchy's criterion $f$ has a right limit at $x_0$. A similar argument shows that $x_0$ has a left limit and that $a$ has a right limit, $b$ a left limit.

Conversely, if $f$ is regulated, we fix $\varepsilon>0$. Then for all $x\in [a,b]$, we can find $\eta(x)>0$ such that if $y,z\in (x-\eta(x),x)\cup (x,x+\eta(x))$ we have $|f(x)-f(y)|<\varepsilon$. Let $I_n:=[a,b]\cap (x-\eta(x),x+\eta(x))$. Then $(I_x)_{x\in [a,b]}$ is an open cover of $[a;b]$, so exists a $\eta>0$ such that for all open interval $I\subset [a,b]$ of diameter $<\eta$ is contained in a set $I_x$. Let $(t_0,\ldots,t_m)$ a subdivision of $[a,b]$ such that $\max_it_{i+1}-t_i<\eta$, $x_i$ such that $(t_i,t_{i+1})\subset I_{x_i}$. Let $S=(a_1,\ldots,a_p)$ a subdivision containing $t_i$ and $x_i$. We have for all $y,z\in (a_i,a_{i+1})$: $|f(y)-f(z)|<\varepsilon$. Fix $c_j:=f\left(\frac{a_i+a_{i+1}}2\right)$ and $f_1$ the step function defined by $f_1(x)=c_j$ if $x\in (a_i,a_{i+1})$ and $f_1(a_i)=f(a_i)$. Then $||f-f_1||_{\infty}\leq \varepsilon$.

share|cite|improve this answer
Note that it works for functions with values in a Banach space, since we only used Cauchy criterion. – Davide Giraudo Feb 26 '12 at 22:19
thank you very much – Ilya Feb 26 '12 at 22:23
You're welcome. It's in fact an application of Lebesgue covering theorem. We note that each Borel-measurable function is a pointwise limit of step function, so here we can see that uniform convergence is much more restrictive. – Davide Giraudo Feb 26 '12 at 22:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.