Is there a non-negative regulated function $f:[a,b] \to \mathbb{R}$ such that $f$ is not identically zero on $[a,b]$, but $\int ^b _a f = 0$, and that $f$ is not a step funciton?

A function $f:[a,b] \to \mathbb{R} $ is a regulated function if $\forall \varepsilon > 0 $ there is a step function $ \varphi :[a,b] \to \mathbb{R} $ such that $sup_{x\in[a,b]} |f(x)-\varphi(x)| <\varepsilon$.

A function $g:[a,b] \to \mathbb{R}$ is a step function if there is a finite set of points $P=\{p_i\}_{i=0} ^k$, $P \in [a,b]$ such that $g$ is constant on each open subinterval of $(p_{i-1}, p_i)$.

I think the function $f$ does not exist because $\int ^b _a f = 0$ means the 'area' under $f$ is zero, and since $f$ is non-negative, there can only be a finite number of points that is not zero in order to satisfy the conditions given. However, this would be a step function. Also would it be possible that $f$ has infinitely many number of points that is not zero and still satisfies all the conditions?

I am not sure how to write up a formal proof, or was I completely wrong and the function do exist. Please help me with this problem.

  • $\begingroup$ It depends on whether you also consider $$f(x) = \begin{cases} 1 & \text{if } x=a \\ 0 &\text{otherwise}.\end{cases}$$ a step function. This function is regulated and nonidentically zero, while it's integral is zero. $\endgroup$ – user99914 Nov 8 '15 at 0:23
  • $\begingroup$ @JohnMa I have added the definition for step funciton, so $f$ would be considered as a step funciton. $\endgroup$ – Lucy Nov 8 '15 at 0:36

An example would be $f:[0,1] \to \mathbb R$,

$$f(x) = \begin{cases}\frac 1n &\text{if }x= \frac 1n, n\in \mathbb N,\\ 0 & \text{otherwise}.\end{cases}$$

this function is regulated, not a step function and $\int_0^1 f(x)dx =0$.

A hint: To show that $f$ is regulated, pick $\epsilon >0$. Then there is $N$ so that $\frac 1n < \epsilon$ for all $n\ge N$. Then you can set a step function which is $0$ on $[0,\frac 1N]$ to approximate $f$.

| cite | improve this answer | |
  • $\begingroup$ What is the regulating step function? It is regulated only because the OP's definition probably missed in the definition of a step function the condition that Pi's must be different than each other? or because the sub-intervals are "open"? Thanks in advance for clarifying this point, because it seems to me that the given definition of step function is somehow misleading. $\endgroup$ – A.S.H Nov 8 '15 at 0:59
  • $\begingroup$ Would you please give me a hint on how to prove that $f$ is regulated? $\endgroup$ – Lucy Nov 8 '15 at 1:04
  • $\begingroup$ @Lucy : Please see the edit. $\endgroup$ – user99914 Nov 8 '15 at 1:07
  • $\begingroup$ @A.S.H : The ambiguity, I think, comes from whether or not one consider $\{a\}$ an interval. $\endgroup$ – user99914 Nov 8 '15 at 1:07
  • $\begingroup$ With the correct definition, as now given, a function can be proved to be regulated if and only if it has finite one-sided limits at each point. That is why most of us instantly recognize Mr Ma's example as regulated. If you don't know this characterization then you have a bit of work to do. $\endgroup$ – B. S. Thomson Nov 8 '15 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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