i read this theorem from V.I.Bogachev vol 1 Measure Theory.
A family $\mathcal{F}\subset L_{1}(\mu)$,where the measure $\mu$ takes values in $[0,+\infty]$, is norm bounded in $L_{1}(\mu)$ precisely for every $\mathcal{A}\subset A$ one has $$\sup_{f\in\mathcal{F}}|\int_{A}f d\mu |<\infty $$
Can anyone tell where i can find o simple proof of this theorem, because i can't understand the proof from Bogachev. Thanks.
I wouldn't know about the proof in the book, but here's a proof. It could probably be streamlined some - you should see what it looked like a few days ago. Going to change some of the notation; this is going to be enough typing as it is.
Going to assume we're talking about real-valued functions, so that for every $f$ there exists $E$ with $\left|\int_E f\right|\ge\frac12||f||_1$.
Theorem Suppose $\mu$ is a measure on (some $\sigma$-algebra on) $X$, $S\subset L^1(\mu)$, and $\sup_{f\in S}||f||_1=\infty$. Then there exists a measurable set $E$ with $$\sup_{f\in S}\left|\int_Ef\right|=\infty.$$
Notation: The letter $f$ will alsways refer to an element of $S$; $E$ and $F$ will always be measurable sets (or equivalence classes of measurable sets modulo null sets).
Proof: First we lop a big chunk off the top: Wlog $S$ is countable; hence wlog $\mu$ is $\sigma$-finite. Now we nibble away at the bottom:
Case 1 $\mu$ is finite and non-atomic.
This is the meat of it. It's also the cool part: We imitate the standard proof of the standard uniform boundedness principle, with measurable sets instead of elements of some vector space.
Let $\mathcal A$ be the measure algebra; that is, the algebra of measurable sets modulo null sets. For $E,F\in\mathcal A$ define $$d(E,F)=\mu(E\triangle F)=||\chi_E-\chi_F||_1.$$Now $\mathcal A$ is a complete metric space (complete because it's isometric with a closed subset of $L^1$). Define $$A_n=\{E\in\mathcal A:\sup_f\left|\int_Ef\right|>n\}.$$
$A_n$ is open in $\mathcal A$: Say $E\in A_n$ and choose $f$ so $$\left|\int_Ef\right|-n=\epsilon>0.$$ There exists $\delta>0$ so the integral of $|f|$ over any set of measure less than $\delta$ is less than $\epsilon$. Hence if $d(E,F)<\delta$ we have $$\left|\int_{F} f\right|\ge \left|\int_{E} f\right|-\int_{E\triangle F}|f|>n,$$so $F\in A_n$.
$A_n$ is dense in $\mathcal A$: Say $E\in\mathcal A$ and let $\epsilon>0$. Write $$X=\bigcup_{j=1}^NE_j,$$where $E_j\cap E_k=\emptyset$ and $$\mu(E_j)<\epsilon.$$ Choose $f$ with $||f||_1>4Nn$. Choose $F$ so $$\left|\int_{F} f\right|>2Nn.$$Now there exists $j$ with$$\left|\int_{F\cap E_j} f\right|>2n.$$We want to show there exists $E'\in A_n$ with $d(E,E')<\epsilon$. If $\left|\int_{E\setminus(F\cap E_j)} f\right|>n$ then $E'=E\setminus(F\cap E_j)$ works; if not the triangle inequality shows that $E'=E\cup(F\cap E_j)$ works.
So the Baire category theorem shows that $\bigcap A_n\ne\emptyset$.
Case 2 $\mu$ is non-atomic. If there exists $E$ with $\mu(E)<\infty$ and $\sup_f\int_E|f|=\infty$ we're done by Case 1. Suppose not.
Suppose we've chosen $f_1,,\dots f_n$ and $E_n$ so that $\mu(E_n)<\infty$ and $$\left|\int_{E_n} f_j\right|>j\quad(1\le j\le n).$$There exists $F$ with $\mu(F)<\infty$, $E_n\subset F$, and such that if $E_{n+1}$ is any set with $$E_{n+1}\cap F=E_n$$then we will still have $$\left|\int_{E_{n+1}} f_j\right|>j\quad(1\le j\le n).$$Choose $c$ so $\int_F|f|\le c$ for all $f$. Choose $f_{n+1}$ with $||f_{n+1}||_1>3c+2(n+1)$. Then $\int_{X\setminus F}|f_{n+1}|>2c+2(n+1)$, so there exists $F_n\subset X\setminus F$ with $\left|\int_{F_n} f_{n+1}\right|>c+{n+1}.$If we let $E_{n+1}=E_n\cup F_n$ then we have $$\left|\int_{E_{n+1}} f_j\right|>j\quad(1\le j\le n+1).$$(For $1\le j\le n$ this follows by the comments above and for $j=n+1$ it uses the triangle inequality.)
So if $E=\bigcup E_n$ then $$\left|\int_{E} f_j\right|\ge j$$for all $j$.
Case 3 $X$ is a countable union of atoms. We may as well assume $\mu$ is a measure on $\Bbb N$. The argument in this case is really just like the argument in Case 2; details available on request.
Case 4 $\mu$ is $\sigma$-finite. Write $X=A_2\cup A_3$ where $\mu$ is non-atomic on $A_2$ and $A_3$ is a countable union of atoms. There exists $j=2,3$ such that $\int_{A_j}|f|$ is unbounded; we are done by Case $j$ above.