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Let $f_n : [0,1]\to \mathbb R$ be a sequence of Riemann integrable functions. We call $f$ Riemann integrable if $U(f) = \inf_P U(f,P) = \sup_P L(f,P) = L(f)$ where $U(f,P)$ denotes the upper sum with respect to partition $P$ and $L(f,P)$ the lower sum.

If $f_n \to 0$ pointwise then it is possible to produce an example of $f_n$ such that $\int f_n$ is unbounded. It is even possible with $f_n$ continuous. But I want to show that it's not possible anymore if $f_n$ are uniformly bounded (I'm quite sure this is true). I tried to prove my conjecture but failed and I'm not sure what the problem is.

Let $f_n \to 0$ pointwise and let $|f_n|\le M$ for some $M > 0$. Let $\varepsilon > 0$. Then the goal is to show that for $n$ greater some $N$:

$$ \int_0^1 f_n \le \varepsilon$$

The definition of Riemann integral lets us use either upper or lower sum instead: If for all partitions $P$ we have $U(f,P) \le \varepsilon$ then $U(f) \le \varepsilon$. But this doesn't seem to be helpful here since we have no information about $f_n$.

How to prove that if $f_n \to 0$ pointwise and $|f_n|\le M$ then $\int f_n \to 0$?

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    $\begingroup$ See this for an elementary proof (i.e., no measure theory). $\endgroup$ – David Mitra Jul 8 '14 at 19:22
  • $\begingroup$ @DavidMitra Thank you. This fully answers my question. Can you post your comment as an answer? $\endgroup$ – blue Jul 9 '14 at 7:12
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This is difficult to do without using measure theory! The result is known as Arzela's Bounded Convergence Theorem.

An elementary proof proof can be found here.

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  • $\begingroup$ Here is an updated link. $\endgroup$ – David Mitra Oct 25 '19 at 14:24
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It is straightforward from the Lebesgue Dominated Convergence Theorem and the fact that any Riemannian integrable function is a Lebesgue integrable function on $[0,1]$.

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