Such questions should be really asked on AoPS rather than here, but, once you've already posted it on MO, I'll answer.
1) The set of zero measure can always be ignored when performing Lebesgue integration, so to say $g_n\to 0$ everywhere or almost everywhere is practically the same: just drop the measure zero set where the convergence fails and apply the bounded convergence theorem as you know it to the integral over the rest.
2) Yes, "uniformly bounded" means here that there is one bound for all functions simultaneously. In this context there is any difference between saying "uniformly bounded sequence" and "bounded sequence" but there is a clear difference between saying "a sequence of bounded functions" and "a sequence of uniformly bounded functions".
Choose $\varepsilon > 0$ and consider sets defined by
$$\Sigma_k = \{E:\ \left|\int\limits_{E} (f_n-f_m) \right| \leqslant \varepsilon, \textrm{ if } n,m \geqslant k \}$$
Since for any measurable set a limit of integrals exists, we have
$\Sigma = \bigcup\limits_{k} \Sigma_k$. Note, that given an integrable function $f$, the functional $f(E):= \int\limits_{E}f$ is continuous respect to $E$ in metric $\delta$. Indeed, $f(E)-f(F) = \int f \cdot ( 1_E - 1_F )$, hence $|f(E)-f(F)| \leqslant \int |f| \cdot | 1_E - 1_F | = \int |f| \cdot 1_{E \Delta F} = \int\limits_{E\Delta F} |f|$.
The last expression tends to $0$ if $|E\Delta F|$ tends to $0$ because of integrability of $f$, equivalently if $d(E,F)\to 0$. This remark shows that sets $\Sigma_k$ are closed as an intersection of closed subsets of the space $(\Sigma,d)$.
From Baire theorem we obtain that one of sets $\Sigma_k$ has an interior point. Therefore, there exists a measurable set $E_0$ and integer $k$ such that the inequality
$ |f_n(E)-f_m(E) | \leqslant \varepsilon$ holds, whenever $|E\Delta E_0| \leqslant \delta$ and $m,n\geqslant k$. We will show, that this inequality holds in fact for any set $E$, provided that its measure is sufficiency small.
By identities $\mathbf{1}_{E\cup E_0} - \mathbf{1}_{E_0} = \mathbf{1}_{E\cap E_0^{c}}$ and $\mathbf{1}_{E_0}-\mathbf{1}_{E_0\setminus E} = \mathbf{1}_{E\cap E_0}$ we obtain for an arbitrary integrable $f$
$$f(E) = f(E \cap E_0^{c}) + f(E\cap E_0) = f(E\cup E_0) - f(E_0) + f(E_0) - f(E_0\setminus E)$$
If $|E| < \delta$, then all of sets $E_0,E_0\cup E, E_0\setminus E$, belong to the ball $\{E:\ |E\Delta E_0| < \delta \}$. Applying the last inequality to $f_n-f_m$ and $|E| < \delta$ we get
$$ |f_n(E)-f_m(E)| \leqslant 2\varepsilon \quad \textrm { if } |E|<\delta, \ n,m\geqslant k$$
Finally, observe that the finite family of integrable functions $f_1,\ldots, f_k$ is obviously locally uniformly integrable, i.e. $\sup\limits_{i\leqslant k} |f_i(E)| \to 0$ if $|E|\to 0$. Therefore, for sufficiency small $\delta'$ we have
$$|f_i(E)|\leqslant \varepsilon \quad \textrm{ if } |E| < \delta', i\leqslant k$$
Gluing together two estimates that have been derived, we see that for some positive $\delta$
$$ \left|\int\limits_{E} f_n\right| \leqslant 3\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
We have estimated integrals for functions $f_n$, instead their modulus. It doesn't matter, however. Namely, applying the last estimate to the set $E\cap \{f_n > 0\}$ and $E\cap \{f_n < 0\}$ respectively (both contained in $E$ hence with a smaller measure), gives finally
$$ \int\limits_{E} |f_n| \leqslant 6\varepsilon, \quad \textrm{ if } |E| \leqslant \delta$$
What finished the proof.
Best Answer
The answer is no. E.g., let $n=1$ and let $B_1$ be the set of all irrational numbers in the interval $[0,1]$. For each $x\in B_1$, let $b_j(x)$ be the $j$th binary digit of $x$. For each natural $j$, let $E_j:=\{x\in B_1\colon b_j(x)=1\}$, so that $\chi_{E_j}=b_j$ on $B_1$.
Then no subsequence of the sequence $(\chi_{E_j})$ converges in measure. Indeed, otherwise there would exist an increasing sequence $(j_k)$ of natural numbers such that the sequence $(b_{j_k})$ converges in measure and hence $b_{j_{k+1}}-b_{j_k}\to0$ in measure (as $k\to\infty$), so that $$\tfrac12=\lambda(\{x\in B_1\colon b_{j_{k+1}}(x)\ne b_{j_k}(x)\}) \\ =\lambda(\{x\in B_1\colon|b_{j_{k+1}}(x)-b_{j_k}(x)|\ge1\})\to0,$$ where $\lambda$ is the Lebesgue measure -- a contradiction.
Thus, no subsequence of the sequence $(\chi_{E_j})$ converges almost everywhere.