# Find a non-negative regulated function that satisfying the given conditions.

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.

• 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. – user99914 Nov 8 '15 at 0:23
• @JohnMa I have added the definition for step funciton, so $f$ would be considered as a step funciton. – 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 $f$.

• 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. – A.S.H Nov 8 '15 at 0:59
• Would you please give me a hint on how to prove that $f$ is regulated? – Lucy Nov 8 '15 at 1:04
• @Lucy : Please see the edit. – user99914 Nov 8 '15 at 1:07
• @A.S.H : The ambiguity, I think, comes from whether or not one consider $\{a\}$ an interval. – user99914 Nov 8 '15 at 1:07
• 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. – B. S. Thomson Nov 8 '15 at 3:39