= Solution
For an elementary <previsible> integrand, the integral is the increment sum $\sum_jH_j(X_{t\wedge t_{j+1}}-X_{t\wedge t_j})$. This expression uses only $H$ and $X$, so is identical under the two measures, irrespective of how the <local martingale> and finite-variation parts change.
We may use the <stochastic dominated convergence theorem>: if bounded <predictable> $H_n$ converge pointwise to $H$ under a common deterministic bound, their integrals against a <semimartingale> converge uniformly on compact intervals in probability. Also convergence in $\mathbb P$-probability implies convergence in $\mathbb Q$-probability when $\mathbb Q\ll\mathbb P$. To see the latter directly, let $D=d\mathbb Q/d\mathbb P$. For any event $E$,
$$
\mathbb Q(E)=\mathbb E_{\mathbb P}(D\mathbf1_E)\le K\mathbb P(E)+\mathbb E_{\mathbb P}(D\mathbf1_{\{D>K\}}),
$$
and let first $\mathbb P(E)\to0$, then $K\to\infty$. Apply this to events involving the supremum on a finite horizon for the uniform-in-probability statement.
Let $\mathcal C$ be the bounded <predictable> integrands for which the two integral processes agree up to $\mathbb Q$-indistinguishability. It is a vector space containing elementary rectangle indicators generating the <predictable> sigma-algebra. It is closed under uniformly bounded pointwise limits: stochastic <dominated convergence theorem> gives convergence of the integrals under each measure, and the preceding transfer makes the $\mathbb P$ limit also a $\mathbb Q$ limit. Uniqueness of limits in probability makes the two limiting integrals equal. The <functional monotone-class theorem> therefore puts every bounded <predictable> integrand in $\mathcal C$.
Finally, for locally bounded $H$, stop on intervals where it has a deterministic bound. Such <stopping times> tend to infinity under $\mathbb P$ and hence also under $\mathbb Q$. Apply the bounded result to the stopped integrands and use the stopping property of the integral. This proves the <stochastic integral under an absolutely continuous measure change> identity
$$
\boxed{(H\cdot X)^{\mathbb P}=(H\cdot X)^{\mathbb Q}\quad\mathbb Q\text{-indistinguishably}.}
$$
The decompositions $H\cdot M+H\cdot A$ and $H\cdot N+H\cdot B$ thus represent the same process under $\mathbb Q$, although their individual components need not agree. With only absolute continuity, equality is asserted outside a $\mathbb Q$-null set, not necessarily outside a $\mathbb P$-null set for a chosen $\mathbb Q$ version.
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