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For an elementary previsible integrand, the integral is the increment sum ${\sum }_j{H}_j(X_{t\wedge t_{j+1}}-X_{t\wedge t_j})$. 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 $E$,and let first $\mathbb{P}(E)\to 0$, 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 identityThe 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.