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I know that if $f_n\geq 0$, that $(f_n)$ is increasing and that if $\lim_{n\to \infty }f_n=f\in L^1$, then, $$\lim_{n\to \infty }\int f_n=\int f.$$

Does this result also work when $(f_n)$ is decreasing ?

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  • $\begingroup$ No, because that would not give the other side of the inequality given by Fatou's lemma. $\endgroup$ – астон вілла олоф мэллбэрг Oct 6 '16 at 12:40
  • $\begingroup$ @астонвіллаолофмэллбэрг: But using dominated convergence, don't we have that $|f_n|\leq f_1\in L^1$ for all $n$ (because $f_n$ decreasing) and thus, using dominated convergence theorem, $\lim_{n\to \infty }\int f_n=\int f$ ? $\endgroup$ – MathBeginner Oct 6 '16 at 13:02
  • $\begingroup$ @MathBeginner You are right. If $f_1 \in L^1$, it is true. You can even take the sequence $f_1-f_n$ and use monotone convergence instead of dominated convergence. $\endgroup$ – Aloizio Macedo Oct 6 '16 at 13:06
  • $\begingroup$ Oh, I understand. Yes, it is true. My mistake earlier, I offer apologies. $\endgroup$ – астон вілла олоф мэллбэрг Oct 6 '16 at 13:08
  • $\begingroup$ @астонвіллаолофмэллбэрг: Great, thank you :-) $\endgroup$ – MathBeginner Oct 6 '16 at 13:08
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If $f_1 \in L^1$ (or any element of the sequence belongs to $L^1$), the dominated convergence theorem (or the monotone convergence theorem applied to $g_n=f_1-f_n$) give you your result.

If not, then it isn't true. For instance, take $f_n=\infty \cdot \chi_ {[0,1/n]}$.

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Let $f_{n}=\frac{1}{n}χ[0,n]$.

For every $\epsilon > 0$ and $x \in \mathbb{R}$ there exists $N > 1$ such that $|f_n(x)| < \epsilon $ for every $n>N$. Hence, $f_n$ converges to $f = 0$ ,uniformly. However, $$0 = \int f d\lambda \not= \lim \int f_n d\lambda = 1$$ The MCT does not apply because the sequence is not monotone increasing. Fatou’s lemma obviously applies.

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  • $\begingroup$ $f_n$ is not decreasing. $\endgroup$ – Aloizio Macedo Oct 6 '16 at 13:19

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