Vtyjkyuk

Let $f$ and $f_n$ measurable numeric functions, $nin mathbb N$ and $f=lim_ntoinftyf_n$ a.e. and suppose an integrable $gge 0 $ exists with $mid f_n mid le g$ a.e. Then $f$ and $f_n$ are integrable with $lim_ntoinftyint mid f_n -f mid dmu=0$.

i) Why is this equivalent to $int fdmu=lim_ntoinfty int f_n dmu$?

ii) And why can one suppose in the proof of this statement (DMCT) that $f$ and $f_n$ are real valued?

Saying $limint|f-f_n|=0$ is not equivalent to saying $limint f_n=int f$ in general. The first implies the second (since $|int f_n-int f|=|int(f_n-f)|leint|f_n-f|$) but not conversely.

Wondering where you could have got the idea that they were equivalent, I think I got it. The following two theorems are equivalent:

Thm 1. Suppose $f_nto f$ almost everywhere, $gge0$, $int g<infty$, and $|f_n|le g$. Then $int f_ntoint f$.

Thm 2. Suppose $f_nto f$ almost everywhere, $gge0$, $int g<infty$, and $|f_n|le g$. Then $int|f_n-f|to0$.

Note that saying the theorems are equivalent does not say that the conclusions of the theorems are equivalent.

It's trivial that Theorem 2 implies Theorem 1. Since I suspect that showing the two theorems are equivalent was a homework problem, I'll just give a huge hint how to show that Theorem 1 implies Theorem 2:

Suppose Theorem 1. Suppose $f_nto f$ almost everywhere, $gge0$, $|f_n|le g$, and $int g<infty$. Let $F_n=???$ and let $G=???$. Then $F_nto F$ almost everywhere, where $F=???$. Also $Gge0$, $|F_n|le G$, and $int G<infty$. So Theorem 1 implies that $int F_ntoint F$, and it follows that $int|f_n-f|to0$.

(That does not show that $int f_ntoint f$ and $int|f_n-f|to0$ are equivalent, because in deriving Theorem 2 from Theorem 1 we applied Theorem 1 to the sequence $(F_n)$, not to the seqence $(f_n)$.)

1)

First, note that

$$left| int f_n - int f right|leq int left|f_n-f right|$$

Thus,

$$lim_nrightarrow inftyint |f_n-f|=0implies lim_nrightarrow inftyint f_n = int f $$

Conversely, note that

$$|f_n-f|leq 2g $$

and that

$$lim_nrightarrow infty |f_n-f|=0$$

Then by DMCT,

$$lim_nrightarrow inftyint|f_n-f| = int 0 = 0 .$$

2) Use that

$$int f = int Re(f) + iint Im(f)$$

and that

$$lim_nrightarrow infty z_n= lim_nrightarrow inftyRe(z_n)+ilim_nrightarrow inftyIm(z_n).$$