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Couple days ago I finished the reading of chapter 6 Riemann-Stiltjes Integral from PMA Rudin. And I noted the he defines Riemann Integral only for bounded functions on $[a,b]$.

But what would be if we consider Riemann integral for $[a,b), (a,b)$ or $(a,b]$? Does sums $U(P,f)$ and $L(P,f)$ changes in these cases? Since we have no information about values $f(a)$ or $f(b)$ because we use them in these above sums.

Let's take a look at this example.

1. Define $f(x)=1$ on $(0,1)$. How to evaluate $\int \limits_{0}^{1}f(x)dx$? In this case we don't nothing about $f(0)$ and $f(1)$. Can anyone show accurately and clearly how it can be evaluated? And prompt how to be in other analogous examples?

I would be really greatful for your help!

RFZ
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3 Answers3

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As long as $f$ is "integrable", you can redefine any finite number of points of $f$ without changing the integral: That is, if $g(x)=f(x)$ everywhere in some interval $I$ except perhaps on the set $\{x_1,x_2,...,x_n\}$, then $\int_Ig(x)dx=\int_If(x)dx$. This fact is not hard to prove directly using Riemann sums.

In particular, if we have $f$ defined only on $(0,1)$, giving values to $f$ at the endpoints will change nothing in evaluating the integral. (More specifically, to evaluate $\int_0^1fdx$ where $f\colon(0,1)\rightarrow\mathbb{R}$ is bounded, just set $f(0)=f(1)=0$)

charlestoncrabb
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  • Dear charlestoncrabb! You wrote really interesting fact that if two functions $f(x)=g(x)$ everywhere except only finite set then then $\int\limits_{I}f(x)dx=\int\limits_{I}g(x)dx$. I tried to prove that before but I have some difficulties. Can you show full proof of this? It would be really great and it would answer to my question. – RFZ Nov 07 '15 at 19:46
  • @RFZ I think you can prove it using Riemann sums, but Im not completely sure now – Masacroso Nov 07 '15 at 19:51
  • The basic idea is to "manage the bad points" $x_1,x_2,...,x_n$ by picking your partition/subintervals so that the subintervals around the bad points are very small so that the contribution to the Riemann sum from these subintervals is very small. – charlestoncrabb Nov 07 '15 at 23:30
  • Also, to make your life easier in manipulating the Riemann sums, try proving instead that $\int_0^1(f-g)dx=0$ – charlestoncrabb Nov 07 '15 at 23:30
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Generally, definite integrals over closed intervals are considered proper and integrals over open or half-open intervals are improper, defined as the limit of proper integrals.

For instance, over the the interval $[0,1)$, the integral is defined as

$$\lim_{x\to 1^+} \int_0^x f(t) \ dt$$

Over the interval open interval $(0,1)$, the integral is defined as a double limit

$$\lim_{x\to 0^-} \lim_{y\to 1^+} \int_x^y f(t) \ dt$$

Simon S
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  • Dear Simon S! Can I claim that your second expression is equal $$\lim_{y\to 1^+}\lim_{x\to 0^-} \int_x^y f(t) \ dt?$$ I just changes limits. – RFZ Nov 07 '15 at 19:41
  • Yes. More carefully we want both of those limits to exist and be equal. Which they will be if one of the limits exists. – Simon S Nov 07 '15 at 19:42
  • So if we define $g=f$ for $x\in (0,1)$, $g(0) =f(0^-), g(1)=f(1^+)$ then the integral over $[0,1]$ of $g$ equals that of your answer? – YoTengoUnLCD Nov 07 '15 at 19:45
  • @YoTengoUnLCD if you mean $g(0) = \lim_{x\to 0^-} f(x)$ if that limit exists, and similarly for $g(1)$, then the integral of $f$ over $(0,1)$ equal the integral of $g$ over $[0,1]$. But in general the values of $g(0), g(1)$ don't matter. – Simon S Nov 07 '15 at 19:48
  • I guess that this claim is true: "if $g(x)=f(x)$ everywhere in some interval $I$ except perhaps on the set ${x_1,x_2,...,x_n}$, then $\int_Ig(x)dx=\int_If(x)dx$" But I never met the proof of it :( – RFZ Nov 07 '15 at 19:52
  • @RFZ That claim is straightforward to prove using the definition of Riemann integration. Using induction, you just need to show adding one point of difference between $g$ and $f$ makes no difference to the integral. – Simon S Nov 07 '15 at 19:53
  • @SimonS, I am gonna prove this interesting fact just now. Can I ask you question if it appear ? – RFZ Nov 07 '15 at 20:04
  • It is equivalent to showing that if

    $$f(x) = \begin{cases} 0, & x \neq a \ c, & x = a \end{cases}$$ for some $a \in [0,1]$ and constant $c$ (without loss of generality $c > 0$), then

    $$\int_I f(x) \ dx = 0$$

    Clearly the lower Riemann sum $L(f,P)$ for all partitions $P$ is zero and you can make the upper sum $U(f,P)$ arbitrarily small.

     – Simon S Nov 07 '15 at 20:07
  • @SimonS, I had the same idea :) But I must to finish it. By the way can I claim that: Let's $g(x)=0$ on $(0,1)$. Taking function $f(x)=0$ on $[0,1]$. Since $f$ and $g$ differs only in two points then $\int fdx=\int gdx$? – RFZ Nov 07 '15 at 20:14
  • The two points are not in the domain of both functions. The two integrals are equal, but only because they both equal zero. What you want to show in your proposition above is that if $f, g$ are both integrable functions $I \to \mathbb R$ which differ at one point $x_D \in I$, then $\int_I f = \int_I g$. To do that, consider $h = f - g$. Then apply induction to show the result for any arbitrary finite set of points of difference ${ x_1, x_2, \cdots, x_n } \subset I$. – Simon S Nov 07 '15 at 20:17
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A Riemann integral over $[a,b]$ has a definition: $$\lim_{\max\Delta x_i\to0}\sum f(x^*_i)\,\Delta x_i$$ where the sum is over the subintervals from some partition of $[a,b]$, with $x^*_i$ coming from a subinterval, and $\Delta x_i$ being the length of that subinterval. The limit is over any sequence of partitions where the maximum subinterval length is eventually arbitrarily small (if you are far enough along in the sequence of partitions). If $f$ is continuous, this limit is guaranteed to exist (meaning, it takes a finite value), by a theorem from calculus.

You could try applying the same definition to an interval with at least one open end. But then the limit is not guaranteed to exist. Examples come in the form of functions that blow up near the open endpoint. For example, $\int\limits_{(0,1]}\frac1x\,dx$ just doesn't exist. They also come in the form of functions with weird behavior near the open endpoint: $\int\limits_{(0,1]}\sin\mathopen{}\left(\frac1x\right)\mathclose{}\,dx$ just doesn't exist either. But as long as $f$ is continuous on $[a,b]$, $\int\limits_{[a,b]}f(x)\,dx$ will always exist.

2'5 9'2
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