Beppo-Levi: Reverse

Question

For a merely decreasing positive sequence it fails: $$f_n:=\frac{1}{n}\chi_{[n,\infty)}:\quad\int f_n\mathrm{d}\lambda=\infty\nrightarrow0$$

For a dominated decreasing positive sequence it holds: $$0\leq f_n\downarrow f:\quad\int f_n\mathrm{d}\mu\to \int f\mathrm{d}\mu\quad(f_n\in\mathcal{L})$$ How to prove this via Beppo-Levi?

I'm trying to prove Lebesgue's dominated convergence theorem for positive functions? — C-star-W-star, Sep 21 '14 at 20:35
Can't you just use that $f = (f -s_n) + s_n$, so that $\int f = \int (f -s_n) + \int s_n$. Now use that $\int s_n \to \int f$ and note that you can substract $\int f$ from both sides, because it is finite. — PhoemueX, Sep 21 '14 at 20:50
@PhoemueX: Perfect thats precisely what I needed =D thanks alot!! — C-star-W-star, Sep 21 '14 at 20:58

score 1 · Accepted Answer · edited Dec 12 '14 at 21:56

1

Since my comment seems to answer the question, let me repeat the argument as an answer...

Denote for shorthand: $\bar{f}:=\sup_n f_n$

By measurability for positive functions one has:

$$\int f_n \mathrm{d}\mu = \int \bar{f} \mathrm{d}\mu + \int (\bar{f}-f_n) \mathrm{d}\mu$$ $$\int f \mathrm{d}\mu = \int \bar{f} \mathrm{d}\mu + \int (\bar{f}-f) \mathrm{d}\mu$$

By monotone convergence one has:

$$\int(\bar{f}-f_n)\mathrm{d}\mu\to\int(\bar{f}-f)\mathrm{d}\mu$$

It remains to observe that all integrals are finite to conclude: $$\int f_n\mathrm{d}\mu=\int\bar{f}\mathrm{d}\mu-\int(\bar{f}-f_n)\mathrm{d}\mu\to\int\bar{f}\mathrm{d}\mu-\int(\bar{f}-f)\mathrm{d}\mu=\int f\mathrm{d}\mu$$

edited Dec 12 '14 at 21:56

C-star-W-star

16,275

answered Sep 21 '14 at 21:07

PhoemueX

35,087

@PhoemuX: Do you mind if I adjust your answer? (I slightly modified the problem to the key setting of your answer.) – C-star-W-star Dec 12 '14 at 20:57
As always, feel free to edit :) – PhoemueX Dec 12 '14 at 21:28
@PhoemuX: Thanks. :D – C-star-W-star Dec 12 '14 at 22:24

Beppo-Levi: Reverse

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