definition of "weak convergence in $L^1$" I have encountered two definitions of weak convergence in $L^1$:
1) $X_n\rightarrow X$ weakly in $L_1$ iff $\mathrm{E}(X_n\mathrm{1}_A)\rightarrow \mathrm{E}(X\mathrm{1}_A)$ for every measurable set $A$.
2) $X_n\rightarrow X$ weakly in $L_1$ iff $\mathrm{E}(X_n f)\rightarrow \mathrm{E}(X\mathrm{1}f)$ for every (essentially) bounded measurable function $f$.
my question: are 1) and 2) equivalent?
I see that 2) implies 1) (indicators are bounded), but I have difficulties establishing that 1) implies 2). I tried approximating $f$ by simple functions $f_m$, say, assuming $X_n,X$ are nonnegative for simplicity; the problem: I cannot justify the interchange in the order of taking the limits (first with $n$, and then with $m$). any ideas? I would appreciate any sort of help. many thanks!
 A: Using Vitali-Hahn-Saks theorem or Baire category theorem with $\mathcal F$ endowed with the metric $\rho(A,B)=\mu(A\Delta B)$, we can show for each $\varepsilon>0$, there is $\delta>0$ such that if $\mu(A)\lt \delta$ then $|\mathbb E[X_n\chi_A]|\lt \varepsilon$. Taking $A':=A\cap \{X_n\leqslant 0\}$ and $A'':=A\cap \{X_n\gt 0\}$, we can see that $\mathbb E[|X_n|\chi_A]\lt\varepsilon$ whenever $\mu(A)\lt\delta$. Indeed, for a fixed $\varepsilon\gt 0$, we define $$F_N:=\bigcap_{n\geqslant N}\left\{A\in\mathcal F,\left|\mathbb E\left[X_n\chi_A\right]\right|\leqslant\varepsilon\right\}.$$
Each $F_N$ is closed and $\bigcup_NF_N=\mathcal F$, hence by Baire's theorem, there is $N_0$, $r_0$ and $A_0\in\mathcal F$ such that $B_\rho(A_0,r_0)\subset F_{N_0}$. Let $B$ such that $\mu(B)\lt r_0$. Since $\mu(A_0\Delta (A_0\cup B))\lt r_0$, $\mu(A_0\Delta (A_0\cap B^c))\lt r_0$ and 
$$\int_B X_n\mathrm d\mu=\int_{A_0\cup B}X_n\mathrm d\mu-\int_{A_0\cap B^c}X_n\mathrm d\mu,$$
we have $\left|\int_B X_n\mathrm d\mu\right|\leqslant 2\varepsilon$ whenever $n\geqslant N_0$ and $\mu(B)\lt r_0$.
Now we use Theorem 1.12.9 in Bogachev, Measure theory, volume 1: 

Let $(\Omega,\mathcal F,\mu)$ be a measure space with a finite non-negative measure. Then for each $\delta>0$, we can find an integer $N$ and a finite partition of $\Omega$, $\{S_1,\dots,S_N\}$ such that for each $i$, either $\mu(S_i)\leqslant \delta$ or $S_i$ is an atom of measure $>\delta$.

So take $\varepsilon:=1$, the associated $\delta$, and notice that there are only finitely many atoms of measure $\gt \delta$. On each of these atoms, $X_n$ is constant.  
