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

After some clarification Clarification On Exercise 7 of Chapter 6 of Baby Rudin.

My attempt: To prove: If $f \in \mathscr{R}$ on $[a,b]$, then $ \int_0^1 f(x) \ \mathrm{d} x = \lim_{c \to 0} \int_c^1 f(x) \ \mathrm{d} x $. Since $f \in \mathscr{R}$ on $[0,1]$, $f \in \mathscr{R}$ on $[0,c]$ and $[c,1]$, $\forall c \in [0,1]$. Let $F(c)= \int_c^1 f(x) \ \mathrm{d} x $, for $c \in [0,1]$. By theorem 6.20, $F(c)$ is continuous on $c \in [0,1]$. Since it (domain of $F$) is closed, it contains all of it’s limit point. By theorem 4.6, $\lim_{c \to 0} F(c)= \lim_{c \to 0} \int_c^1 f \ \mathrm{d} x = F(0)= \int_0^1 f \ \mathrm{d} x$. Now by definition 4.25, $F(0)=F(0+)= \lim_{c \to 0^+} \int_c^1 f \ \mathrm{d} x= \int_0^1 f \ \mathrm{d} x$. Our desired equality.

Is this proof correct?

user264745
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  • By definition, to prove the equality you shall show that to every $\epsilon>0$ corresponds a $\delta>0$ such that $\lvert\int^1_0-\int^1_c\rvert<\epsilon$ if $0<c<\delta$. –  Dec 14 '21 at 05:38
  • You cannot apply theorem 6.20, since rudin didn't define $\int^a_a=0$, $\int^b_a=-\int^a_b$. –  Dec 14 '21 at 05:57
  • @user912011 we can prove those properties, don’t we? – user264745 Dec 14 '21 at 07:21
  • No. See definition 6.1. –  Dec 14 '21 at 09:00
  • Usually, one defines $\int^b_a=-\int^a_b$ since $\int_a^b+\int_b^c=\int_a^c$ and $\int_a^a=0$, $a<b<c$. –  Dec 14 '21 at 09:02
  • @user912011 In theorem 6.20, Rudin defines $F$ for all $[a,b]$. So Rudin is implicitly assuming all these properties of integral. – user264745 Dec 14 '21 at 09:23
  • No. In proof of Theorem 6.20, Rudin avoids talking about $\int^b_a$ where $a\geq b$. –  Dec 14 '21 at 10:00
  • In fact $F'$ is a one sided derivative –  Dec 14 '21 at 10:01
  • @user912011 Domain of $F$ is $[a,b]$. So $F(a)=\int_a^a f(t) dt$. – user264745 Dec 14 '21 at 10:07
  • @user912011 By definition of function, $\int_a^a$ should be define. – user264745 Dec 14 '21 at 10:08
  • @user264745 I solved your question I just need to type it, maybe tomorrow. – Mr. Proof Dec 19 '21 at 15:24
  • What a coincidence! I come up with this solution independently. Now I found out that in this https://math.stackexchange.com/q/2375688/861687 post, in the comment section(last comment by Paramanand Singh) there is a hint to this approach(using theorem 6.20). – user264745 Dec 20 '21 at 17:26

1 Answers1

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A) Suppose $f \in \mathcal{R}$ on $[0,1]$. Let $\epsilon >0$. Let $M= \sup\{ |f(x)|: 0 \leq x \leq 1\}$. Fix $c \in (0, \frac{\epsilon}{4M})$, and consider any partition of $[0,1]$ that contains $c$, where the lower and upper Riemann sums are $\sum M_j (t_j - t_{j-1})$ and $\sum m_j (t_j - t_{j-1})$ of $f$ differ by less than $\frac \epsilon 4$. Then the partition of $[c,1]$ formed by the points of this partition that lie in the interval surely has the property that its upper and lower Riemann sum $\sum M'_j(t_j - t_{j-1}) $ and $\sum m'_j (t_j - t_{j-1})$ differ by less than $\frac \epsilon 4$. From above we have

$$\sum M_j (t_j - t_{j-1}) \frac \epsilon 4 < \int_0^1 f(x) dx < \sum m_j (t_j - t_{j-1}) + \frac \epsilon 4$$ and

$$\sum M'_j (t_j - t_{j-1}) \frac \epsilon 4 < \int_c^1 f(x) dx < \sum m'_j (t_j - t_{j-1}) + \frac \epsilon 4$$

Moreover, we have

$$\left|\sum M_j (t_j - t_{j-1}) - \sum M'_j (t_j - t_{j-1}) \right| < \frac \epsilon 4$$

and

$$\left|\sum m_j (t_j - t_{j-1}) - \sum m'_j (t_j - t_{j-1}) \right| < \frac \epsilon 4.$$

From these two inequalities you can see that

$$\left|\int_0^1 f(x) dx - \int_c^1 f(x)dx \right|< \epsilon.$$

B)- let

$$f(x) = (-1)^n (n+1).$$

For $\frac{1}{n+1} < x\leq \frac{1}{n}, n=1,2,...$ Then if $\frac{1}{N+1} \leq \leq \frac{1}{N}$ we have

$$\int_c^1 f(x) dx = (-1)^N (N+1) (\frac 1 N -c) + \sum_{k=1}^{k-1} \frac{(-1)^k}{k}.$$

Since $0\leq \frac{1}{N} -c \leq \frac{1}{N} - \frac{1}{N+1} = \frac{1}{N (N+1)}$, the first term on the right-hand side tends to zero as c goes to zero, while the sum converges to $\ln 2$. However,

$$\int_c^1 |f(x)| dx = (N+1) (\frac{1}{N}) + \sum_{k=1}^{N-1} \frac{1}{k},$$ although in this case the first term in the right-hand side goes to zero, in contrast, the summation goes to infinity.

José Carlos Santos
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Mr. Proof
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    This answer seems different approach to the problem. Before I look at this solution, I will try first myself. So it will take some time. – user264745 Dec 20 '21 at 12:33
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    @user264745 When you work on mathematical analysis you have to prove everything analytically that’s why the solution should look like rigorous a little bit. – Mr. Proof Dec 20 '21 at 12:34
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    Yeah I completely agree. – user264745 Dec 20 '21 at 12:36
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    Hey! first of all thank you so much for showing interest to this post. I first thought your answer to (a) would be similar to this one https://math.stackexchange.com/q/2375688/861687. But it isn’t. So after reading first few lines of proof(to know what direction to go). I wrote the following proof: After some observations. $\forall \epsilon \gt 0$, take $\delta:= \epsilon / M \gt 0$ such that if $0 \lt c \lt \delta$(since it’s a one sided limit), then $| \int_0^1 f dx - \int_c^1 f dx|=|\int_0^c f dx| \leq Mc \lt M\delta=\epsilon$. Is this proof correct? – user264745 Dec 20 '21 at 17:14
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    Because the references I cited, OP did something more complicated(in taking the $\delta$). – user264745 Dec 20 '21 at 17:14
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    I have read your answer (a). This is maybe third way to solve this problem(Ex 6.7). It is little bit complicated. It will take some time to absorb/familiar with all the notation. – user264745 Dec 20 '21 at 17:35
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    Potential errors: (1)$c\in (0,\epsilon /4M)$. What if $\epsilon/4M \gt 1$? So possibility of $1 \lt c \lt \epsilon /4M$. (2)for given $\epsilon \gt 0$, take partition $P$ of $[0,1]$, s.t. $c\in P$ & $U(P,f)-L(P,f) \lt \epsilon /4$. Is there rigours way to prove $\exists P$ s.t. $c\in P$ & $U(P,f)-L(P,f) \lt \epsilon /4$? If the constraint $c \in P$ weren’t there, then it’s trivially true. (3)$U(P,f) \epsilon /4 \lt \int f dx \lt L(P,f)+ \epsilon /4$. why is there multiplication of $\epsilon /4$ factor in both $U(P,f)$ and $U(P_c,f_c)$, where $P_c=P \cap [c,1]$ & $f_c$ defined on $[c,1]$? – user264745 Dec 21 '21 at 18:43
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    (4)$U(P,f) \epsilon /4 \lt \int_0^1 f dx \lt L(P,f)+ \epsilon /4$. As far as I know, $L(P,f) \leq \int_0^1 f dx \leq U(P,f) \lt U(P,f)+\epsilon /4$. We may very well have $\epsilon /4 \gt 1$ & $U(P,f)\gt 0$, in that case $U(P,f)\lt U(P,f)\epsilon /4 \lt \int_0^1 f dx$, which is a straight contradiction. (5) How $U(P,f)-U(P_c,f_c) \lt \epsilon /4$? (6) How did you obtain last desired inequality? – user264745 Dec 21 '21 at 18:51
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    @user264745 There is no error. The solution is perfect. – Mr. Proof Dec 24 '21 at 01:59
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    Can you please explain each of the points, I am having doubt with? – user264745 Dec 24 '21 at 07:11
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    I really don’t understand your proof, especially multiplication of $\epsilon /4$ in $U(P,f)$. – user264745 Dec 25 '21 at 18:10
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    @user264745 I think you have some confusion about the understanding of the $\epsilon$. First, we assume epsilon is arbitrary. For instance, you can choose $epsilon = \frac 1 k$ and you can find a partition, in your problem, that satisfies the inequality, being able to choose smaller (decreasing and bounded below) epsilons so the sup of epsilon will be zero. Multiplying epsilon by any fixed value is just a matter of making the solution clear and neet. – Mr. Proof Dec 26 '21 at 00:22
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    @user264745 I don’t know why you didn’t mark my answer as the right answer! – Mr. Proof Dec 28 '21 at 02:57
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    Ohhh... I’m sorry. I totally forgot about this whole post situation. Now I’m studying topology, and it keeping me busy. – user264745 Dec 28 '21 at 06:45
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    I’m sorry I didn’t understand your solution “completely”. To reply your sentence of your last comment. Multiplying epsilon by any fixed value have the potential of changing the inequality sign. Is there any way you can acknowledge each point(of 6 Que) I raised as my doubt? I know it might be frustrating to you that I don’t understand the solution. I’m sorry for that. – user264745 Dec 28 '21 at 06:53
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    @user264745 I meant multiplying the epsilon by any positive value (expected to be large and finite). The situation is that if we want to prove (mathematically) that two values become closer as an index (or discrete or continuous-time variable) goes to infinity we, in principle, use the epsilon (delta of large N) argument. How? For instance in sequences when we want to prove that $x_n \to x$ in $\mathbb{R}$ we prove that $\forall \epsilon >$ $\exists N>0$ such that

    $$\forall n > N \Rightarrow |x_n -x| <\epsilon.$$

    Here there is implicit information about epsilon. Shown below:

     – Mr. Proof Dec 28 '21 at 09:21
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    @user264745 The challenge is to find a large enough $N>$ for all smaller and smaller epsilon. Here we are assuming doing this process, pick epsilon and find good N, infinitely many times. Now let's define the set that contains all epsilons: $$A_1= { \frac 1 k; k \in \mathbb{N}}$$, so every time we choose an epsilon, say $\frac 1 k$ and find suitable N, which depends on epsilon and the limit x in pointwise convergence $N=N(\epsilon,x)$ and on just epsilon in uniform convergence, the next time we choose smaller epsilon $\frac{1}{k+1}$. Since $$\inf A_1 =0$$ then we find eventually $x_n \to x$ – Mr. Proof Dec 28 '21 at 09:31
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    @user264745 Now is the time to answer your question; define $$A_M = { \frac{M}{k}| k \in \mathbb{N}, 0< M< \infty},$$ you can clearly see that $\inf { A_M} =0$ for every positive and finite M. I hope these comments answers your questions. – Mr. Proof Dec 28 '21 at 09:33