A question on proving countable additivity

[zzzhhh]

This question comes from the proof of Lemma 9.3 of Bartle's "The Elements of Integration and Lebesgue Measure" in page 97-98. This proof is shown as the image below.

Form (9.1) mentioned in the lemma is: $$(a,b], (-\infty,b], (a,+\infty), (-\infty,+\infty)$$.

My question is: although $$I_j$$ constructed in P98 is a bit fatter than $$(a_j,b_j]$$, I doubt the assertion that the left endpoint a, and in turn the compact interval [a,b], is also covered by $$\{I_j\}$$, as the proof in the text claimed (I drew a red underline). Is my doubt correct (this means the text is incorrect), or point a can be proved to be covered by $$\{I_j\}$$ (how)? Thanks!
PS: the establishment of the converse inequality does not need the coverage of the whole [a,b]. A small shrink, say $$[a+\epsilon,b]$$, is sufficient to get the inequality.

 

[Office_Shredder]

Definitely seems to be an error in the text. If $$a_i=a+\frac{1}{n}$$ and $$\epsilon=\frac{1}{2}$$ and $$\epsilon_i=\frac{1}{2^{i+1}}$$ the $$I_j$$ never include $$a$$

It seems true that you can pick to $$\epsilon_i$$'s so that you get a covering... we know that the $$a_i$$ have to get arbitrarily close to $$a$$, so you can pick one really close to $$a$$ to add one of your larger values of $$\epsilon_i$$ to

 

[zzzhhh]

Thank you Office_Shredder!

 

[DrRocket]

This question comes from the proof of Lemma 9.3 of Bartle's "The Elements of Integration and Lebesgue Measure" in page 97-98. This proof is shown as the image below.

Form (9.1) mentioned in the lemma is: $$(a,b], (-\infty,b], (a,+\infty), (-\infty,+\infty)$$.

My question is: although $$I_j$$ constructed in P98 is a bit fatter than $$(a_j,b_j]$$, I doubt the assertion that the left endpoint a, and in turn the compact interval [a,b], is also covered by $$\{I_j\}$$, as the proof in the text claimed (I drew a red underline). Is my doubt correct (this means the text is incorrect), or point a can be proved to be covered by $$\{I_j\}$$ (how)? Thanks!
PS: the establishment of the converse inequality does not need the coverage of the whole [a,b]. A small shrink, say $$[a+\epsilon,b]$$, is sufficient to get the inequality.

I have not gone through the entire argument, but the condition 9.2 and the ordering of the a's and b's assumed would require that $$a=a_1$$ and $$b_i=a_{i+1}$$.