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If a bounded function $f$ is not defined on a finite set $A$ of numbers inside an interval $[a,b]$, is it Riemann integrable over that interval $[a,b]$?

I have seen that some people jump over this question by defining the function in an arbitrary way, $0$ for example, on the set of numbers where it is not defined. But this seems wrong to me, since then you are modifying the domain, and therefore integrating a different function, not the original one.

My confusion increases, because it seems to me that the $f$ might be Darboux integrable, without the need of defining the function in the set $A$. But I have read that Riemann and Darboux are equivalent.

Any help would be welcome. Thanks in advance.

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  • $\begingroup$ A bounded (but not continuous) function on say $[0,1]$ need not be Riemann integrable. It won't help if, in addition, it is not bounded at a few points. $\endgroup$
    – Mirko
    Commented Mar 16, 2021 at 2:30
  • $\begingroup$ Changing the definition of a function at a finite number of points does not matter as far as its Riemann integrability or the value of its integral is concerned. Hence there is no problem if you define the function at a finite number of points in any manner. However you can't do so for an infinite number of points. $\endgroup$
    – Paramanand Singh
    Commented Mar 16, 2021 at 4:34
  • $\begingroup$ This is very convenient in dealing with integrals like $\int_0^1\sin(1/x)\,dx$. One just defines the integrand in any manner at $x=0$. Some people take another approach of defining it as $\lim_{t\to 0^+}\int_t^1\sin(1/x)\,dx$ but this is unnecessary and cumbersome. We use the improper integrals when either 1)the function is unbounded or 2) the domain is unbounded. $\endgroup$
    – Paramanand Singh
    Commented Mar 16, 2021 at 4:39
  • $\begingroup$ See related math.stackexchange.com/q/3694790/72031 $\endgroup$
    – Paramanand Singh
    Commented Mar 16, 2021 at 4:41

1 Answer 1

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First of all, it seems like you might be thinking that all bounded functions are Riemann integrable, so I want to disavow you of that notion. Take $f(x) = 1$ for $x$ rational and $0$ for $x$ irrational. This is bounded on all compact intervals but is not Riemann integrable on any.

Strictly speaking, the definition of a Riemann integral is on a bounded function $f: [a, b] \longrightarrow \mathbb R$, so $f$ must be defined on all of $[a, b]$ no ifs ands or buts. If we had some $f$ defined on $[a, b] - A$ then it is technically incorrect to write $\int_a^b f(x) dx$. In fact, we cannot even ask if $f$ is Riemann integrable on $[a, b]$ as it is not a real function defined on $[a, b]$.

However, we are not always so strict in math. We can choose to abuse the notation of the Riemann integral and write $\int_a^b f(x) dx$ even when $f$ is not defined everywhere on $[a, b]$, but we must justify this abuse. Indeed, let $g_1, g_2: [a, b] \longrightarrow \mathbb R$ such that $g_1$ is Riemann integrable and $g_2$ differs from $g_1$ at only finitely many points. Then $g_2$ is Riemann integrable and $\int_a^b g_1(x) dx = \int_a^b g_2(x) dx$. To see why, we can work with the difference $h = g_1 - g_2$ and prove that $h$ is Riemann integrable with integral $0$. Roughly speaking, as $h$ is nonzero at only finitely many points, shrinking our partition around these points allows our Riemann sums to tend to $0$.

What this fact about $g_1, g_2$ says is that the Riemann integral does not care about the values of your function on any finite set. This is why we feel justified in talking about the Riemann integrability of a partially defined function $f: [a, b] - A \longrightarrow \mathbb R$. Whatever extension to the whole of $[a, b]$ we choose makes no difference - if $g_1, g_2: [a, b] \longrightarrow \mathbb R$ both extend $f$ then $g_1$ is Riemann integrable iff $g_2$ is, in which case their integrals coincide. We can therefore comfortably say that $f$ is Riemann integrable, when we mean that an extension is, and we can comfortably compute its integral by computing the integral of any extension.

To put this a bit more formally, let $R([a, b])$ be the set of all Riemann integrable functions on $[a, b]$. This is actually a real vector space and we have a linear map $\int_a^b: R([a, b]) \longrightarrow \mathbb R$ sending $f \mapsto \int_a^b f(x) dx$. Now, we define an equivalence relation on $R([a, b])$ via $f \sim g$ if $f(x) = g(x)$ at all but finitely many points. The quotient $R([a, b])/{\sim}$ is a real vector space, and the integral descends to a well defined map $R([a, b])/{\sim} \longrightarrow \mathbb R$ sending $[f] \mapsto \int_a^b f(x) dx$. This is just formalism for my above statement that the Riemann integral doesn't care about the values on a finite set. Additionally, take a partially defined function $f: [a, b] - A \longrightarrow \mathbb R$ where $A$ is finite. Suppose that $f$ can be extended to a Riemann integrable function on $[a, b]$. Then there is no canonical choice of $g \in R([a, b])$ to assign to $f$. However, there is a canonical equivalence class to assign to $f$ - that is, $[g]$ where $g$ is any extension of $f$. In fact, we can feel justified referring to this equivalence class as just $[f]$. Point being, this object $[f]$ is still specific enough to talk about Riemann integrability as much as we want. As an aside, this equivalence relation idea is extremely fruitful when dealing with the Lebesgue integral, where we replace "finite" with "measure 0."

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  • $\begingroup$ (+1) Nice answer. I was thinking about giving a brief comment along the lines of your 2nd paragraph with a nod to what you say in your 3rd paragraph, but this is much better than anything I was thinking of saying. $\endgroup$ Commented Mar 16, 2021 at 7:15
  • $\begingroup$ That's very nice of you to say, thank you! $\endgroup$ Commented Mar 16, 2021 at 7:50
  • $\begingroup$ Thanks for your answer my friend. $\endgroup$ Commented Mar 17, 2021 at 17:52
  • $\begingroup$ Glad I could help! $\endgroup$ Commented Mar 17, 2021 at 18:02

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