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Currently I am reading Rick Durrett's Probability: Theory and Examples. I am going through the exercises of ch 1.5 I have trouble with the following exercise (ex 1.5.5)

If $ g_n \uparrow g $ (pointwise) and $\int g_1^{-} \,d\mu< \infty $ then $\int g_n\, d\mu \uparrow \int g\, d\mu$

My plan is to prove that $ g_n^- \uparrow g^- $ and by monotone convergence $\int g_n^- d\mu \uparrow \int g^- d\mu$ the same could be done for $ g^+_n $ and since $g = g^+ - g^-$ the exercise is (maybe) done.

The part that confuses me is the $\int g_1^{-} < \infty $ part, which says that $ g_1^{-} $ is Lebesgue integrable. Here are my questions:

  • Why is this needed in the exercise ?
  • The monotone convergence theorem requires the functions to be greater then zero but not integrable, why is that ? (not related to the exercise)
  • Are functions that are greater then zero all integrable, since one step of the construction of the integral is the construction of the integral for functions greater then zero ? (not related to the exercise)

PS.

  • notation is taken from the book : $ g^{-} = max\{-g, 0\}$
  • by "function greater then zero" I mean $ f(x) \geq 0 \space for\space all \space x $
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  • $\begingroup$ Do you mean g1 is the first term of the sequence ??? $\endgroup$
    – Devendra Singh Rana
    Sep 10, 2017 at 19:01
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    $\begingroup$ I retyped the exercise from the book (one to one ) and no explanation is given there. I supposed that g1 is the first element of the sequence. What else could it mean ? $\endgroup$
    – Arseny Nerinovsky
    Sep 10, 2017 at 19:30
  • $\begingroup$ Ya probably it's the same thing, or may be it's a typo becoz a condition on only the first term doesn't really makes sense $\endgroup$
    – Devendra Singh Rana
    Sep 11, 2017 at 2:50

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