= Solution
The <Radon-Nikodym theorem> for positive measures says: if $\mu$ and $\nu$ are <sigma-finite measures> on the same <measurable space>, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative <measurable function> $g$, unique $\mu$-almost everywhere, such that
$$
\nu(E)=\int_E g\,d\mu\qquad(E\text{ measurable}).
$$
Here <sigma-finiteness> means that the space is a countable union of measurable sets of finite measure; <absolute continuity of measures>, written $\nu\ll\mu$, means that $\mu(E)=0$ implies $\nu(E)=0$. The function $g=d\nu/d\mu$ is the <Radon-Nikodym derivative>. For a finite signed or <complex measure> $\nu$ of finite <total variation norm of a measure>, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu)$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.
\b[For every measure space and $1<p<\infty$, $(L^p)^*$ is isometrically $L^q$, where $1/p+1/q=1$.] We use the complex-linear pairing
$$
\Phi_g(h)=\int_\Omega hg\,d\mu.
$$
With the convention $\int h\overline g$, the same identification is conjugate-linear in $g$. By <Hölder's inequality>, $\Phi_g$ is a <bounded linear functional> with $\|\Phi_g\|\le\|g\|_q$. If $g\ne0$, set
$$
h=\frac{\overline g\,|g|^{q-2}}{\|g\|_q^{q-1}},
$$
interpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\|h\|_p=1$ and $\Phi_g(h)=\|g\|_q$. Thus
$$
\boxed{\|\Phi_g\|=\|g\|_q.}
$$
It remains to represent an arbitrary $\Phi\in(L^p)^*$, rather than merely produce functionals from $L^q$.
We prove <Lp duality on an arbitrary measure space> without imposing sigma-finiteness on $\mu$. Write $C=\|\Phi\|$. For every <measurable set> $E$ with $\mu(E)<\infty$, define a <complex measure> on $E$ by
$$
\nu_E(F)=\Phi(\mathbf1_F)\qquad(F\subseteq E\text{ measurable}).
$$
It is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their <indicator functions> tend in $L^p$ to $\mathbf1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu|_E$, since <indicator functions> of <null sets> represent zero in $L^p$.
Its <total variation norm of a measure> is finite. For any finite measurable partition $E=\bigcup_j E_j$, choose scalars $c_j$ of modulus $1$ with $c_j\nu_E(E_j)=|\nu_E(E_j)|$. Then
$$
\sum_j|\nu_E(E_j)|=\Phi\left(\sum_jc_j\mathbf1_{E_j}\right)
\le C\left\|\sum_jc_j\mathbf1_{E_j}\right\|_p
=C\mu(E)^{1/p}.
$$
Taking the supremum over partitions gives the variation bound. Since $\mu|_E$ is finite, the <Radon-Nikodym theorem> supplies $g_E\in L^1(E)$ with $\nu_E(F)=\int_F g_E\,d\mu$. By uniform approximation with <simple functions>,
$$
\Phi(h)=\int_E hg_E\,d\mu
$$
for every bounded measurable $h$ supported in $E$.
To improve $g_E$ from $L^1$ to $L^q$, test with the bounded function
$$
h_N=\overline{g_E}|g_E|^{q-2}\mathbf1_{\{|g_E|\le N\}}\mathbf1_E.
$$
If $I_N=\int_{E\cap\{|g_E|\le N\}}|g_E|^q\,d\mu$, then
$$
I_N=\Phi(h_N)\le C\|h_N\|_p=C I_N^{1/p}.
$$
Thus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The <monotone convergence theorem> gives
$$
\int_E|g_E|^q\,d\mu\le C^q.
$$
If $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the <Radon-Nikodym theorem>.
We now perform <support localization of an Lp functional>. Set
$$
M=\sup_{\mu(E)<\infty}\int_E|g_E|^q\,d\mu\le C^q.
$$
Choose <finite-measure sets> $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup\cdots\cup E_n$, $S=\bigcup_nD_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a <measurable function> $g$ on $S$ by taking $g_{D_n}$ on the disjoint <measurable sets> $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,
$$
\int_\Omega|g|^q\,d\mu=\lim_n\int_{D_n}|g_{D_n}|^q\,d\mu=M.
$$
Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.
For any <finite-measure set> $F\subseteq\Omega\setminus S$, compatibility on the disjoint union $D_n\cup F$ gives
$$
\int_{D_n}|g_{D_n}|^q\,d\mu+\int_F|g_F|^q\,d\mu\le M.
$$
Letting $n\to\infty$ forces $g_F=0$ almost everywhere. For an arbitrary <finite-measure set> $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi(h)=\int hg\,d\mu$ for every <simple function> supported on a <finite-measure set>.
Those <simple functions> are dense in $L^p$ even for this arbitrary <measure space>. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\|h\|_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by <simple functions>; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and <Hölder's inequality> therefore extend the representation to every $h\in L^p$.
We have constructed $g\in L^q$ with $\Phi=\Phi_g$, and the previously proved norm identity gives $\|g\|_q=\|\Phi\|$. It also proves uniqueness: if $\Phi_g=\Phi_{\widetilde g}$, then $\|g-\widetilde g\|_q=0$. This completes the isometric <duality of Lp spaces>.
\b[The dominated sequence actually converges to zero in norm, and hence weakly.] The assumptions give $|f_n|^p\le|f|^p\in L^1$ and $|f_n|^p\to0$ almost everywhere. The <dominated convergence theorem> yields
$$
\|f_n\|_p^p=\int|f_n|^p\,d\mu\longrightarrow0.
$$
For every $\Phi\in(L^p)^*$, $|\Phi(f_n)|\le\|\Phi\|\|f_n\|_p\to0$. Therefore
$$
\boxed{f_n\xrightarrow{w}0.}
$$
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