4

I am sure the answer to this question is already somewhere here, but I could not find it.

If we have a sequence of functions $f_n \in L_1$, with pointwise limit $f$, and assuming we know that $\int |f_n| d\mu \rightarrow \int |f| d\mu$, where the latter might possibly be infinite, does this imply $\int f_n d\mu \rightarrow \int f d\mu$?

The reason I am asking this is because I was confused if I could apply monotone convergence in the case that I do not have monotonicity of the functions itself but rather of the absolute values, i.e. $|f_n|\leq |f_{n+1}|$? And if not, is there a good counterexample that illustrates how this approach will fail?

a_student
  • 1,504

2 Answers2

1

I believe Scheffé's Lemma answer's your question affirmatively. It says that if $f_n \to f$ a.e. where $f_n, f\in L^1$ then $$ \int |f_n -f|d\mu \to 0 \iff \int |f_n|d\mu \to \int |f|d\mu$$

E G
  • 991
  • 2
    Minor quibble: under OP's assumptions, it might happen that $f \notin L^1$. For example, consider $\mathbb{N}$ with the counting measure, and take $f(k) = (-1)^k$ and $$ f_n(k) = \begin{cases} (-1)^k,&\text{if $k \leq n$};\ 0,&\text{otherwise}. \end{cases} $$ – Josse van Dobben de Bruyn May 23 '21 at 18:16
  • @JossevanDobbendeBruyn Good point thank you for mentioning that! – E G May 23 '21 at 19:00
  • @Oliver Diaz thanks for the counterexample! If you would make this an answer, I would accept it (since it also shows how the mentioned application of monotone convergence could fail). (but @E G , this answer is of course also very helpful) – a_student May 24 '21 at 14:03
  • @a_student: I'm glad you found my comment helpful. I just posted it as an answer at your request. – Mittens May 24 '21 at 14:30
  • Another remark. In the OP could it happen that $f_n \to f$ a.e. and $\int |f_n|$ converges, but not to $\int |f|$ ?? – GEdgar May 24 '21 at 16:16
1

IF $|f|=\infty$ the answer may be no. For example $$f_n(x)=\sum^n_{j=1}(-1)^j j\mathbb{1}_{(j,j+1]}(x)$$ $f_n\xrightarrow{n\rightarrow\infty}f:=\sum^\infty_{j=1}(-1)^j j\mathbb{1}_{(j,j+1]}(x)$, $\lim_n\int |f_n|=\infty=\int|f|$, and $\lim_n\int|f-f_n|=\infty$.

When $f\in L_1$, the answer is yes and it follows from the following version of dominated convergence

If $f_n\in L_1$, $g_n\in L_1$, $f_n\xrightarrow{n\rightarrow\infty}f$ $\mu$-almost surely, $|f_n|\leq g_n$, and $\lim_n\int g_n=\int\lim_n g_n$, then $$\int|f-f_n|d\mu\xrightarrow{n\rightarrow\infty}0$$ In particular $\lim_n\int f_n\,d\mu=\int f\,d\mu$.

by setting $g_n=|f_n|$.

Edit: For a discussion of the version of dominated convergence I quoted above you may see this.

Mittens
  • 39,145