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We say that $f$ is regulated on $[a,b]$ if there is a sequence of step function $(f_n)$ on $[a,b]$ s.t. $f_n \to f$ uniformly.

Does this statement hold : $f:[a,b]\to \mathbb R$ is Riemann integrable on $[a,b]$ $\iff$ $f$ is regulated.


If $f$ is regulated, it's obviously Riemann integrable.

Q1) But what happen for the implication ? Let $f$ integrable, then of course $\lim_{n\to \infty }\sum_{i=0}^{n-1}M_i\boldsymbol 1_{[x_i,x_{i+1}]}=f(x)$ for all $x\in [a,b]$ where $M_i=\max_{[x_i,x_{i+1}]}f$, but I have difficulties to prove that the convergence is uniform.

Q2) If instead of $[a,b]$ we have $[0,\infty )$ does the statement is still true ? For example, let $f$ a regulated function on $[0,\infty )$, i.e. there is a sequence of step functions $(f_n)$ that converge uniformly to $f$. Now, $$\lim_{n\to \infty }\int_0^\infty f_n=\lim_{n\to \infty }\lim_{M\to \infty }\int_0^M f_n(x)dx,$$ and I dont really see why we can permute $\lim_{n\to \infty }$ and $\lim_{M\to \infty }$.

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The fact that regulated functions have one-sided limits leads to an easy counterexample:

The function $f(x)=\begin{cases} \sin\frac{1}{x} & 0<x<1\\ 0 & x=0 \end{cases}$

is Riemann integrable on $[0,1]$ but it is not regulated.

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  • $\begingroup$ $\sin(1/x)$ is really Riemann integrable on $[0,1]$ ? $\endgroup$ – user657324 Mar 30 at 15:30
  • $\begingroup$ Yes, because it has at most countably many points of discontinuity. (It has only one in fact). $\endgroup$ – Matematleta Mar 30 at 15:32
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    $\begingroup$ Very Nice, indeed :) $\endgroup$ – user657324 Mar 30 at 15:33
  • $\begingroup$ Thanks! The point is that the class of regulated functions is more restrictive because the convergence of the step functions must be $uniform.$ $\endgroup$ – Matematleta Mar 30 at 15:39
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Q1 has been answered, with a simpler counterexample than the one I had in mind. For Q2: Define $f:[0,\infty)\to[0,1]$ by $f(t)=1$. Then $f$ iis regulated but not (improper) Riemann integrable on $[0,\infty)$.

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