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I want to find examples for

a) A sequence $(f_n)_{n\in \mathbb N}$ of non-negative, measurable functions that converge pointwise, but $$\int f_n \,d\mu$$ does not converge.
b) A sequence $(f_n)_{n\in \mathbb N}$ of measurable functions with $f_1 \leq f_2 \leq \dotsc$ with existing pointwise limit $f:= \lim_{n\to \infty} f_n$ but $$\int f_n \, d\mu \neq \int f\, d\mu$$

Attempt: a)
Define $f_n := \left ( 1 + \frac{1}{n}\right ) \mathbb 1_{[0,n]}$. Then $f_n \to 1$ pointwise but $$\int f_n(x) \, dx= n+1 \to \infty$$ Is this correct? Moreover, is it possible to find a function like this but we allow convergence in $\mathbb R \cup \{ \infty \}$, i.e. where the sequence of integrals must not have a limit in $\mathbb R \cup \{ \infty \}$?
b)
Define $f_n = -\frac{1}{n} \mathbb 1_{]-\infty, \frac{1}{n}]}$. Then $f_n \to 0$ pointwise, but $$\int f_n(x) \, dx = -\infty \neq 0 = \int 0 \, dx$$ again, is this correct and does there exists a solution with finite values? Thanks in advance!

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  • $\begingroup$ Your example for (a) is fine. To find an example where the sequence of integrals has no limit in the extended reals, consider the sequence $g_n = \mathbb 1_{(0,1/n)}$. This converges pointwise to zero. Now consider multiplying $g_n$ by an appropriate sequence of coefficients to achieve your goal. For example, $f_n = n^2 g_n$ integrates to $n$, hence the sequence of integrals diverges to $+\infty$. Can you modify this so that you get some other type of divergence? $\endgroup$ – Bungo Apr 17 '18 at 14:58
  • $\begingroup$ How about I define $h_n := n$ if $n$ is even and $h_n := 2n$ if $n$ is odd. Then $f_n := h_n g_n$ does the job? :) And how about (b)? Thanks for your time! $\endgroup$ – Staki42 Apr 17 '18 at 16:43
  • $\begingroup$ Yes, that works. Your answer for (b) is also correct. There is no solution with finite values; see this answer for details. $\endgroup$ – Bungo Apr 18 '18 at 3:32

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