Riemann integral and Lebesgue integral (in real analysis Folland)

Question

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

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)

Answer 1

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.