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(1) A function $f:[a,b]\to\mathbb{R}$ is said to be Riemann integrable on $[a,b]$ . . .

(2) A function $f:(a,b)\to\mathbb{R}$ is said to be Riemann integrable on $(a,b)$ . . .

Suppose a book started the exposition like (1) and other started like (2). My question is: Has every theorem that is valid for first definition, a analogue for the second?

I think most of the books (at least my calculus books and analysis books) uses closed intervals, but I started to study Lebesgue Integral and in the first chapter of book the author presents a review about Riemann integral and considers functions defined on open intervals.

Maybe it's trivial, but I would like know if there is a reason for choose to work with open intervals instead of closed intervals.

Thanks.

Michael Hardy
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Pedro
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1 Answers1

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There is no such thing as a "Riemann integral defined over the open interval $\ ]a,b[\ $".

The definition of the Riemann integral deals with bounded functions defined on a compact ($=$ finite closed) interval $[a,b]$.

It turns out that when you take a function which is Riemann-integrable on $[a,b]$ and change its values on a very small set, e.g., on a finite set, then the value of the integral does not change. It follows that when you have a function on the open interval $\ ]a,b[\ $ which is Riemann-integrable on $[a,b]$ for any values you might choose at $a$ and $b$ then the Riemann integral of $f$ over $[a,b]$ is well defined.

It is a completely different matter when you are forced to consider integrals of the form $$\int_{a+\epsilon}^{b-\epsilon'} f(x)\ dx$$ with $\epsilon$, $\epsilon'>0$ in order to get hold of the "integral" you actually want, e.g., when you want to consider the integral $$\int_0^1{1\over\sqrt{x(1-x)}}\ dx\ .$$ In this case one would speak of an improper Riemann integral.

Christian Blatter
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