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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.

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  • $\begingroup$ 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
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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 $f$.

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  • $\begingroup$ 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

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