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 arguement, 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}.