# Mistake in Bartle's proof of Hake's Theorem?

Here is Bartle's proof of Hake's Theorem found in "A Modern Theory of Integration". I think there is a mistake in the highlighted line:

The Theorem: $f:[a,b]\to \mathbb{R}$ is gauge integrable if and only if it is gauge integrable in $[a,x]\ \forall x\in [a,b)$ and the limit $\lim_{x\to b}\int_a^xf$ is finite. In that case $$\int_a^bf=\lim_{x\to b^-}\int_a^xf$$

The problem is in the $(\impliedby)$ proof:

Let $(c_n)$ be a strictly increasing sequence with $c_0=a$, $c_n\to b$. As $\lim_{n\to +\infty}\int_{c_0}^{c_n}f=\lim_{x\to b^-}\int_a^xf=L\in \mathbb{R}$ and $c_n\to b$, $\exists r\in \mathbb{N}$ so that $$\forall x\in (c_r,b)\ \left|\int_a^xf-L\right|<\epsilon\text{ and }\left|f(b)\right|(b-c_r)<\epsilon$$ By the integrability in $[c_{n-1},c_n]$, there exists a gauge $\delta_n$ on $[c_{n-1},c_n]$ so that if $\dot{\mathcal{P}}_n<<\delta_n$, $$\left|S(f,\dot{\mathcal{P}}_n)-\int_{c_{n-1}}^{c_n}f\right|<\frac{\epsilon}{2^n}$$ Without loss of generality, \begin{align*}\delta_1(c_0)&\le \frac12(c_1-c_0)\tag{i}\\ \delta_{n+1}(c_n)&\le \min\left\{\delta_n(c_n),\frac12(c_n-c_{n-1}),\frac12(c_{n+1}-c_n)\right\}\tag{ii}\\ \delta_n(x)&\le \min\left\{\frac12(x-c_{n-1}),\frac12(c_n-x)\right\}\text{ for }x\in (c_{n-1},c_n)\tag{iii} \end{align*} We define a gauge in $[a,b]$ by $$\delta(x)=\begin{cases}\delta_n(x)&\text{ if, }x\in [c_n,c_{n+1})\\ b-c_N&\text{ if, }x=b\end{cases}$$ Let $\mathcal{P}=\left\{a=x_0<...<x_m=b\right\}$ partition $[a,b]$ with tags $t_i$ so that $\dot{\mathcal{P}}<<\delta$. As $b\not\in\cup_{n=0}^{\infty}[c_n,c_{n+1})=[a,b)$, this forces $b$ to be a tag of $[x_{m-1},b]$. But then, as $t_m=b$, $$c_r=b-\delta(b)<x_{m-1}$$ Let $s\in \mathbb{N}$ be the smallest integer with $x_{m-1}\le c_s$ so that $r\le s$. If $k=1,...,s-1$ then condition (iii) implies that the point $c_k$ must be a tag of the sub-interval in $\mathcal{P}$ that contains it. ...

My argument: Suppose $c_k\in [x_p,x_{p+1}]$. But, $[x_p,x_{p+1}]$ may also contain other points of the sequence (finitely many). Let $c_{q},c_{q+1},...,c_k,c_{k+1},...,c_l$ be all these points. Suppose $t_p$ is the tag of $[x_p,x_{p+1}]$. If $t_p\neq c_q,c_{q+1},...,c_l$ and $t_p>c_k$ then $t_p\in (c_i,c_{i+1})$ with $c_i\ge c_k$. Condition (iii) and the fact that $\dot{\mathcal{P}}<<\delta$ imply that $$c_k> t_p-\delta(t_p)=t_p-\delta_i(t_p)\ge t_p-\frac{t_p-c_i}2=\frac{t_p+c_i}2\ge \frac{c_k+c_k}2=c_k$$ which is a contradiction. Similarly if $t_p<c_k$. What I have shown is that the tag must be a point of the sequence. But why must it be $c_k$ like Bartle says?. This fact is crucial to the proof. Bartle then continues:

Using the right-left procedure we may assume $c_0,c_1,...,c_{s-1}\in \mathcal{P}$.

That can't be done (I think). Indeed consider the simple case where the only points of the sequence in $[x_p,x_{p+1}]$ are $c_k$ and $c_{k+1}$. The tag $t_p$ must be one of $c_k,c_{k+1}$. Suppose it were $c_k$. Then $c_k\in \mathcal{P}$ and $c_{k+1}\in (c_k,x_{p+1})$. How is it then possible to prove $c_{k+1}$ is the tag of $[c_k,x_{p+1}]$?. $c_k$ seems like an equally strong candidate for this.

EDIT: I also think there is a gap in "As $b\not\in\cup_{n=0}^{\infty}[c_n,c_{n+1})=[a,b)$, this forces $b$ to be a tag of $[x_{m-1},b]$." Why can't a point $c_n$ be the tag?

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