The following is from p.57(Real Analysis, by Folland)

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My question is following:

  1. Dominated convergence theorem should roughly be: $f_n \rightarrow f, |f_n|\leq g$ a.e., then $\int f = \text{lim}\int f_n$ (not detailed version). How to understand the following is from dominated convergence theorem (there is no $f_n$):

    $$\int_{[0,\infty]} f \ \ dm=\text{lim}_{b\rightarrow \infty} \int_0^b f(x) \ \ dx$$

  2. Next, it says "even when $f$ is not integrable". Is this "not Riemann integrable"? And what does that example tell us? (hope for detailed explanation)

  1. You can express the limit in terms of sequences. For any sequence $b_n\to\infty$, the $f_n$ in this case would be $f(x)\chi_{[0,b_n]}$ and the dominated convergence theorem would yield $\int f_n\to\int f$.

  2. It means "Lebesgue integrable". The function is not Lebesgue integrable because the integrals of the positive and negative parts would evaluate to $\sum \frac 1 {2k} =\infty$ and $\sum \frac 1{2k+1} = \infty$, respectively. On the other hand, the improper Riemann integral evaluates to the alternating series $\sum (-1)^k\frac 1 k$, which converges to a finite value.

    What this example tells us is that the Lebesgue integral is able to generalize the proper Riemann integrals but only a subset of the improper Riemann integrals. It does so by showing how the Lebesgue integral can fail to give a result that an improper integral gives.


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