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Q1) We often defined the Riemann integral of a function with Darboux sum, but could we define the space of Riemann integrable function as the closure of step functions ? (but in "$L^1$"-sense) (as Lebesgue Integrable function are defined as the closure of simple function). I mean, $f$ is a Riemann integrable function on $[a,b]$ $\iff$ there is a sequence a sequence of step functions s.t. $$\lim_{n\to \infty }\int_a^b f_n=\int_a^b f$$where $f$ are step functions.

Q2) I have an other small question (but related to the previous one). Regulated function are Riemann integrable, but is the converse also true ? I.e. all Riemann integrable function are Regulated. I recall that a regulated function is a uniform limit of step function.

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  • $\begingroup$ What would be the meaning of $\int_a^b f$ in this definition? Is $f$ assumed Lebesgue-integrable? $\endgroup$ – punctured dusk May 3 '18 at 15:26
  • $\begingroup$ As there is a sequence of step function $(f_n)$ s.t. $\lim_{n\to \infty }\int_a^b f_n$ $\endgroup$ – MathBeginner May 3 '18 at 15:27
  • $\begingroup$ It gives you Lebesgue integrable again. $\endgroup$ – user551819 May 3 '18 at 15:29
  • $\begingroup$ @totoro: of course not ! $\endgroup$ – Surb May 3 '18 at 15:45
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    $\begingroup$ Tone down the rhetoric. Civility is required in the comments. $\endgroup$ – Alexander Gruber May 3 '18 at 15:58
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First, just saying that $$\int_a^b f_n\to\int_a^b f$$does not imply that $f_n\to f$ in any interesting or reasonable sense. For example, if $f$ is Riemann integrable you could just let each $f_n$ be the constant function $$f_n=\frac1{b-a}\int_a^b f.$$It seems more reasonable to ask this: If $f$ is Riemann integrable does there exist a sequence of step functions $f_n$ with $$\int_a^b|f-f_n|\to0?$$The answer to that is yes; this follows from the characterization of the Riemann integral in terms of Darboux sums. Hint: If $m_j\le f\le M_j$ on the interval $I_j$ then $|f-m_j|\le M_j-m_j$ on $I_j$.

Can this be used to give an alternate definition of the Riemann integral? I don't see how - if we don't know what the integral is then we don't have a definition for $\int_a^b|f-f_n|$.

No, a Riemann integrable function need not be regulated. Because for example a regulated function has one-sided limits at every point.

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  • $\begingroup$ Thanks a lot, I think I see better. Could you please answer for Q2) ? $\endgroup$ – MathBeginner May 3 '18 at 15:59
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For Q1), the fact that simple function are dense in Lebesgue integrable functions is a consequence, not a definition ! Normally, $f$ is integrable in lebesgue sense if $$\sup\left\{\int_a^b \varphi\mid \varphi\leq f\text{ simple}\right\}=\inf\left\{\int_a^b \varphi\mid f\leq \varphi\text{ simple}\right\},$$ and we write $\int_a^b f$ this number. For Riemann, we say that $f$ is Riemann integrable if $$\sup\left\{\int_a^b \varphi\mid \varphi\leq f\text{ step function}\right\}=\inf\left\{\int_a^b \varphi\mid f\leq \varphi\text{ step function}\right\}.$$

But to define $\int_a^b f$ as "there is a sequence of step function $(f_n)_n$ s.t. $$\int_a^b f=\lim_{n\to \infty }\int_a^bf_n$$ has unfortunately no sense. May be you mean that : "if there is a sequence $(f_n)_n$ s.t. $f_n(x)\to f(x)$ for all $x\in [a,b]$, then we define $\int_a^b f$ as $$\int_a^b f=\lim_{n\to \infty }\int_a^b f_n,$$ but unfortunately, you can have an other sequence of step function $(g_n)$ s.t. $g_n(x)\to f(x)$ for all $x\in [a,b]$" but unfortunately it can happen that$$\lim_{n\to \infty }\int_a^b f_n\neq \lim_{n\to \infty }\int_a^b g_n,$$ and thus $\int_a^bf$ wouldn't be well defined (since depending on the sequence).

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  • $\begingroup$ Thanks a lot, I think I see better. Could you please answer for Q2) ? $\endgroup$ – MathBeginner May 3 '18 at 15:58
  • $\begingroup$ I think that your definition of when a function is Lebesgue integrable only works for bounded functions ? Simple functions are by definition real (or complex) rather than extended real en.wikipedia.org/wiki/Simple_function. $\endgroup$ – Tom Collinge Mar 1 at 13:37
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For Q2, not every Riemann integrable function is regulated.

A simple counterexample is $f:[0,1] \to [0,1]$ where

$$f(x) = \begin{cases}1, \quad x = 1/2^k, \,\, k \in \mathbb{N}\\ 0, \quad \text{otherwise} \end{cases}$$

Clearly, $f$ is Riemann integrable since the set of discontinuities has measure zero.

Consider a partition $(x_0,x_1,\ldots, x_n)$ of $[0,1]$ and a step function $\phi(x) = \sum_{j=1}^n a_j \mathbf{1}_{(x_{j-1},x_j]}.$

If $f$ were regulated there would exist a step function $\phi$ such that $|f(x) -\phi(x)| < 1/4$ for all $x \in [0,1]$. Thus $|f(x) - a_1| < 1/4$ for all $x \in [0,x_1]$.

This is impossible since $f$ takes both the values $0$ and $1$ at infinitely many points in $[0,x_1].$ We cannot have both $|0 - a_1| = |a_1| < 1/4$ and $|1 - a_1| < 1/4$, since $1 - |a_1| < |1 - a_1| < 1/4 $, which implies $|a_1| > 3/4$.

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  • $\begingroup$ Neat, but I would rather say that "for any choice of the partition point $x_1$ the function takes both values $0$ and $1$ in $(0, x_1]$ " ... etc. $\endgroup$ – Tom Collinge Feb 28 at 14:18

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