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I'm reading the following theorem in Folland:

Let $f$ be a bounded real-valued function on $[a,b]$. If $f$ is Riemann integrable, then $f$ is Lebesgue measurable (and hence integrable on $[a,b]$ since it is bounded), and $\int^b_a f(x)\,dx = \int_{[a,b]}f \, dm$.

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It's not clear to me how Dominated Convergence Theorem is being used to conclude $$\int G \, dm = \int g \, dm = \int_a^bf(x)\,dx$$ What's dominating? Where's the sequence? Also they conclude that $G=f$ a.e. after showing $G=g$ a.e. How did they get to $G=f$ a.e.?

EDIT: $$S_Pf = \sum_1^nM_j(t_j-t_{j-1}) \quad s_Pf = \sum_1^nm_j(t_j-t_{j-1})$$ where $M_j$ and $m_j$ are max and min on a piece of the partition

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  • $\begingroup$ Perhaps you should define the notation in the proof (e.g. $S_P$, etc). Other users can infer what they mean, but it’s probably best to avoid ambiguity. Also, I think there’s a typo in the theorem you quote. (What is $B$?) $\endgroup$ – Theoretical Economist Sep 25 '18 at 20:45
  • $\begingroup$ Just a side note: the result that you are working on only states that if a function is Riemman integrable, it is Lebesgue integrable and the integrals coincide. The other implication is false (take for example the characteristic function of the rationals in that interval), and so this is not really an equivalence. $\endgroup$ – Guido A. Sep 25 '18 at 20:49
  • $\begingroup$ @TheoreticalEconomist sorry fixed, I've just been staring at this for a while, I forgot stuff was defined earlier $\endgroup$ – yoshi Sep 25 '18 at 20:58
  • $\begingroup$ @GuidoA. well the conclusion follows only if the hypothesis holds - namely $f$ is Riemann integrable. though I agree, perhaps its abusing notation a bit. $\endgroup$ – yoshi Sep 25 '18 at 21:00
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    $\begingroup$ $f$ is Riemann integrable,so you can take $any$ sequence of partitions whose mesh $\to0$, to calculate $\mathscr R\int f$. So, choose nested partitions $P_k$ with the property that mesh $\to 0$ as $k\to \infty$ and the fact that they are nested means that the $G_{P_k}$ are decreasing, so that the $\lim G_{P_k}$ exists. In particular, $|G_{P_1}|$ is Lebesgue integrable and $f\le |G_{P_1}|$ so the DCT may be applied. I personally think the proof in Baby Rudin is much clearer, with the added benefit of getting also the fact that $f$ is Riemann integrable iff it's continuous a.e.$\lambda.$ $\endgroup$ – Matematleta Sep 25 '18 at 21:21

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