Contradiction to Continuity of Integration?

Question

For each of the two functions $f$ on $[1,\infty)$ defined below, show that $\lim_{n \rightarrow \infty} \int_1^n f$ exists while $f$ is not integrable over $[1,\infty)$. Does this contradict the continuity theory of integration?

(i) Define $f(x) = (-1)^{-n} / n$, for $n \leq x < n +1 $.

(ii) Define $f(x) = (\sin x )/x$ for $1 \leq x < \infty $.

This is Royden 4e pg 91. I think the idea is that with the Lebesgue integral we are calculating the positive and negative parts separately and then combining, and in this case their improper Riemann integral would cancel, but in this theory it does not work. How do I go about showing this. Do I investigate $\mathbb{R}\setminus(\infty, 1)$?

Answer 1

These functions are not Lebesgue-integrable since their absolute value isn't (definition of lebesgue-integrability for real-valued functions).

For the first function this is easy ($\int |f(x)|dx = \sum_{n} \frac{1}{n} = \infty$).

For the second function we have $|f(n\pi + x)| > \frac{1}{\pi}\frac{\sin(x)}{n+1}$ for $n>1$ and $x \in [0,\pi[$ so we have (roughly) $\int |f(x)|dx > \frac{1}{\pi} \times \int_{0}^{\pi} \sin(x)dx \times \sum_{n} \frac{1}{n} = \infty$.

However, you can integrate over $[1,n]$ and take the limit, for the first you get $\sum_{n} \frac{(-1)^{n}}{n} = \ln(2)$ and for the second you can integrate by parts.

The continuity of integration is not contradicted since it assumes the function to be integrable, which is not here.

Answer 2

If by $\mathbb R\setminus(\infty,1)$ you meant $\mathbb R\setminus\{\infty,1\}$, then the answer is that deleting two points won't help because the Lebesgue integral is not altered by changing the value of a function at a finite set of points. And $\infty$ is not a member of $\mathbb R$.

I don't know what Royden means by "continuity of integration". The most celebrated theorem that might go by that name is the dominated convergence theorem, which says $$ \int_A \lim_n f_n = \lim_n \int_A f_n\quad\text{ if }\quad\int_A \sup_n |f_n| <\infty. $$ This is a moderately big "if". There are cases in which the equality fails despite the fact that the expressions on both sides of the equality are well defined, and of course in those cases one has $\int_A \sup_n |f_n|=\infty$.

The two examples you cite are cases in which equality fails because the expression on one side of the equality is undefined, i.e. $\displaystyle\int_{(1,\infty)}\frac{\sin x}{x}\,dx$ is undefined because $\displaystyle\int_{(1,\infty)}\left|\frac{\sin x}{x}\right|\,dx=\infty$. Whether that "violates continuity of integration" would have to depend on how "continuity of integration" is defined.