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(Baby Rudin Chapter 6 Exercise 7a)

Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c, 1]$ for every $c>0$. Define \begin{equation}\tag{1} \int_0^1 f(x) \, dx = \lim_{c \to 0} \int_{c}^1 f(x) \, dx \end{equation} if this limit exists (and is finite). If $f \in \mathscr{R}$ on $[0, 1]$, show that this definition of the integral agrees with the old one.

I wrote a proof for this question for $0<c<1$ and I am wondering how to deal with the cases when $c=0$ or $c=1$. Firstly, is either (or both) of these cases even realized? I am guessing that $c=0$ is never realized.

If $c=1$, then I think that $f \in \mathscr{R}$ on $\{1\}$ and consequently, $\int_1^1 f(x) \, dx =0$. Moreover, $\lim_{c \to 0} \int_{c}^1 f(x) \, dx = \lim_{c \to 0} \int_{1}^1 f(x) \, dx =0$ and $(1)$ is trivial. Is this argument correct?

Edit: The (only) answer to this question seems to assume that $0<c<1$.

Ricky_Nelson
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  • The question is about a limit as $c\to0$, so of course only values of $c$ near $0$ are relevant. – Andreas Blass Aug 11 '20 at 16:37
  • @AndreasBlass Does this mean that $c=0$ is invalid? What about $c=1$? Is $1$ close enough to $c$? – Ricky_Nelson Aug 11 '20 at 16:52
  • $c=0$ is irrelevant to the right side of the given equation. Also, as I said before, only values of $c$ near 0 are relevant. Check the definition of $\lim_{c\to0}$. – Andreas Blass Aug 11 '20 at 16:57

1 Answers1

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Actually, asserting that you have a proof for $0<c<1$ doesn't make sense. The question assumes that the restriction of $f$ to $[c,1]$ is Riemann-integrable for every $c>0$. If, say, you only assume that $f$ is Riemann-integrable on $\left[\frac12,1\right]$, then mentioning the limit$$\lim_{c\to0}\int_c^1f(x)\,\mathrm dx$$makes no sense, since it may well happen that, for instance, the integral $\int_{1/4}^1f(x)\,\mathrm dx$ doesn't exist.

So, the problem is this: assume that $f$ is Riemann-integrable on $[0,1]$ and then prove that$$\int_0^1f(x)\,\mathrm dx=\lim_{c\to0}\int_c^1f(x)\,\mathrm dx.$$

José Carlos Santos
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  • Sorry, but I am not sure what you mean by that the $0<c<1$ part of my proof doesn't make sense. I've added my full proof to my post now. Is it possible that $c>0$ or $c>1$? Your answer makes me think that both $c=0$ and $c=1$ are possible. – Ricky_Nelson Aug 11 '20 at 15:30
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    At no point of my answer I claimed that the $0<c<1$ part of your proof doesn't make sense. What I wrote was that the claim that if $f$ is Riemann integrable on $[c,1]$, for some $0<c<1$, then $\int_0^1f(x),\mathrm dx=\lim_{c\to0}\int_c^1f(x),\mathrm dx$ makes no sense. The problem lies in the statement, not in its proof. – José Carlos Santos Aug 11 '20 at 15:34