[proofplan] We prove Girsanov's theorem in three steps. First, we localize: since $X - \langle X, M \rangle$ is continuous, stopping at the exit time from $[-n, n]$ makes it bounded. It then suffices to show that the bounded stopped process $Y = X^{T_n} - \langle X^{T_n}, M \rangle$ is a $\mathbb{Q}$-martingale. To do this, we show $ZY$ is a $\mathbb{P}$-martingale (where $Z = \mathcal{E}(M)$) by computing $d(ZY)$ via the Itô product rule and observing that the drift terms cancel. Since $Y$ is bounded and $Z$ is uniformly integrable, $ZY$ is uniformly integrable. Finally, we use the Bayes formula for conditional expectations to transfer the $\mathbb{P}$-martingale property of $ZY$ to the $\mathbb{Q}$-martingale property of $Y$. [/proofplan]

[step:Localize $\tilde{X}$ by stopping to reduce to a bounded process] Define $\tilde{X}_t = X_t - \langle X, M \rangle_t$. Since $X$ is a continuous local martingale and $\langle X, M \rangle$ is a continuous finite-variation process, $\tilde{X}$ is a continuous semimartingale. Define the stopping times \begin{align*} T_n = \inf\{t \geq 0 : |\tilde{X}_t| \geq n\}. \end{align*} Since $\tilde{X}$ is continuous, $T_n \uparrow \infty$ $\mathbb{P}$-a.s. Because $\mathbb{Q} \ll \mathbb{P}$ (the Radon-Nikodym derivative $\mathcal{E}(M)_\infty > 0$ $\mathbb{P}$-a.s. since the stochastic exponential of a continuous local martingale is strictly positive), $T_n \uparrow \infty$ $\mathbb{Q}$-a.s. as well. It suffices to show that $\tilde{X}^{T_n}$ is a $\mathbb{Q}$-martingale for each $n$, since this implies $\tilde{X}$ is a $\mathbb{Q}$-local martingale (with reducing sequence $(T_n)$). Set $Y_t = \tilde{X}_{t \wedge T_n} = X_{t \wedge T_n} - \langle X, M \rangle_{t \wedge T_n}$. By the properties of stopped covariation, $\langle X, M \rangle_{t \wedge T_n} = \langle X^{T_n}, M \rangle_t$. Therefore $Y_t = X^{T_n}_t - \langle X^{T_n}, M \rangle_t$, and $|Y_t| \leq n$ for all $t \geq 0$ by the definition of $T_n$. [/step]

[step:Show $ZY$ is a $\mathbb{P}$-martingale via the Itô product rule]Let $Z_t = \mathcal{E}(M)_t$. By the [Stochastic Exponential as a Local Martingale](/theorems/2102) theorem, $Z$ satisfies $dZ_t = Z_t \, dM_t$. Apply the [Integration by Parts](/theorems/2097) formula to the product $Z_t Y_t$: \begin{align*} d(Z_t Y_t) = Y_t \, dZ_t + Z_t \, dY_t + d\langle Y, Z \rangle_t. \end{align*} We compute each term. For $dZ_t$: we have $dZ_t = Z_t \, dM_t$. For $dY_t$: since $Y_t = X^{T_n}_t - \langle X^{T_n}, M \rangle_t$, we have \begin{align*} dY_t = dX^{T_n}_t - d\langle X^{T_n}, M \rangle_t. \end{align*} For $d\langle Y, Z \rangle_t$: since $dZ_t = Z_t \, dM_t$, the covariation of $Z$ with any continuous semimartingale $Y$ satisfies $d\langle Y, Z \rangle_t = Z_t \, d\langle Y, M \rangle_t$. Now $\langle Y, M \rangle = \langle X^{T_n}, M \rangle$ because the finite-variation part $-\langle X^{T_n}, M \rangle$ has zero covariation with $M$. Therefore \begin{align*} d\langle Y, Z \rangle_t = Z_t \, d\langle X^{T_n}, M \rangle_t. \end{align*} Substituting into the product rule: \begin{align*} d(Z_t Y_t) &= Y_t Z_t \, dM_t + Z_t \bigl(dX^{T_n}_t - d\langle X^{T_n}, M \rangle_t\bigr) + Z_t \, d\langle X^{T_n}, M \rangle_t \\ &= Y_t Z_t \, dM_t + Z_t \, dX^{T_n}_t. \end{align*} The $\pm Z_t \, d\langle X^{T_n}, M \rangle_t$ terms cancel exactly. Both $Y_t Z_t \, dM_t$ and $Z_t \, dX^{T_n}_t$ are stochastic integrals against continuous local martingales ($M$ and $X^{T_n}$ respectively) with locally bounded integrands ($Y$ is bounded by $n$, and $Z$ is continuous hence locally bounded). Therefore $ZY$ is a continuous local martingale under $\mathbb{P}$. We now upgrade to a true martingale. Since $Z = \mathcal{E}(M)$ is a uniformly integrable $\mathbb{P}$-martingale by hypothesis, the family $\{Z_\tau : \tau \text{ a stopping time}\}$ is uniformly integrable. Since $|Y_t| \leq n$, the family $\{Z_\tau Y_\tau\}$ satisfies $|Z_\tau Y_\tau| \leq n Z_\tau$, and is therefore uniformly integrable. By the [Characterization of Martingales Among Local Martingales](/theorems/2078), $ZY$ is a $\mathbb{P}$-martingale.[/step]

[guided]The heart of the proof is to compute $d(ZY)$ and observe the cancellation. Let us trace through the product rule carefully. We have two processes: - $Z_t = \mathcal{E}(M)_t$ with $dZ_t = Z_t \, dM_t$ (from the [Stochastic Exponential](/theorems/2102) theorem). - $Y_t = X^{T_n}_t - \langle X^{T_n}, M \rangle_t$ with $dY_t = dX^{T_n}_t - d\langle X^{T_n}, M \rangle_t$. The [Integration by Parts](/theorems/2097) formula for continuous semimartingales states $d(ZY) = Y \, dZ + Z \, dY + d\langle Y, Z \rangle$. **Computing $d\langle Y, Z \rangle$:** The covariation is bilinear and the covariation of a finite-variation process with any semimartingale is zero (by the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem applied to the continuous case). So: \begin{align*} \langle Y, Z \rangle = \langle X^{T_n} - \langle X^{T_n}, M \rangle, Z \rangle = \langle X^{T_n}, Z \rangle, \end{align*} because $\langle X^{T_n}, M \rangle$ is a finite-variation process. Since $dZ_t = Z_t \, dM_t$, the covariation $\langle X^{T_n}, Z \rangle_t = \int_0^t Z_s \, d\langle X^{T_n}, M \rangle_s$ by the bilinearity of covariation and the [Bracket of Two Stochastic Integrals](/theorems/2092) formula. Thus $d\langle Y, Z \rangle_t = Z_t \, d\langle X^{T_n}, M \rangle_t$. **Assembling the product rule:** \begin{align*} d(Z_t Y_t) &= Y_t \, dZ_t + Z_t \, dY_t + d\langle Y, Z \rangle_t \\ &= Y_t Z_t \, dM_t + Z_t(dX^{T_n}_t - d\langle X^{T_n}, M \rangle_t) + Z_t \, d\langle X^{T_n}, M \rangle_t. \end{align*} The $-Z_t \, d\langle X^{T_n}, M \rangle_t$ from $Z \, dY$ and the $+Z_t \, d\langle X^{T_n}, M \rangle_t$ from $d\langle Y, Z \rangle$ cancel, leaving \begin{align*} d(Z_t Y_t) = Y_t Z_t \, dM_t + Z_t \, dX^{T_n}_t. \end{align*} Why do these cancel? Because the $-\langle X^{T_n}, M \rangle$ term in $Y$ was placed there precisely to absorb the covariation between $X^{T_n}$ and $Z$. This is the mechanism of Girsanov: the "drift correction" $-\langle X, M \rangle$ in $\tilde{X} = X - \langle X, M \rangle$ is engineered so that $Z \tilde{X}$ has no finite-variation part, i.e., is a local martingale. Both remaining terms are stochastic integrals against continuous local martingales with locally bounded integrands (since $|Y| \leq n$ and $Z$ is continuous), so $ZY$ is a continuous local martingale under $\mathbb{P}$. **Upgrading to a true martingale:** The hypothesis states $Z = \mathcal{E}(M)$ is a uniformly integrable $\mathbb{P}$-martingale. This means $\{Z_\tau\}$ is uniformly integrable over all stopping times $\tau$. Since $|Y| \leq n$, we have $|Z_\tau Y_\tau| \leq n Z_\tau$ for every stopping time $\tau$, so $\{Z_\tau Y_\tau\}$ is uniformly integrable. By the [Characterization of Martingales Among Local Martingales](/theorems/2078), $ZY$ is a true $\mathbb{P}$-martingale.[/guided]

[step:Transfer the $\mathbb{P}$-martingale property of $ZY$ to the $\mathbb{Q}$-martingale property of $Y$ via the Bayes formula]We use the abstract Bayes formula for conditional expectations. For $s \leq t$ and any bounded $\mathcal{F}_t$-measurable random variable $\xi$, the change of measure formula gives \begin{align*} \mathbb{E}^{\mathbb{Q}}[\xi \mid \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[Z_\infty \xi \mid \mathcal{F}_s]}{\mathbb{E}^{\mathbb{P}}[Z_\infty \mid \mathcal{F}_s]} = \frac{\mathbb{E}^{\mathbb{P}}[Z_\infty \xi \mid \mathcal{F}_s]}{Z_s}, \end{align*} where the last equality uses the $\mathbb{P}$-martingale property of $Z$: $\mathbb{E}^{\mathbb{P}}[Z_\infty \mid \mathcal{F}_s] = Z_s$. Apply this with $\xi = Y_t$. Since $|Y_t| \leq n$, $Y_t$ is bounded. We compute: \begin{align*} \mathbb{E}^{\mathbb{Q}}[Y_t \mid \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[Z_\infty Y_t \mid \mathcal{F}_s]}{Z_s}. \end{align*} By the tower property of conditional expectations applied to the $\mathbb{P}$-martingale $ZY$: \begin{align*} \mathbb{E}^{\mathbb{P}}[Z_\infty Y_t \mid \mathcal{F}_s] = \mathbb{E}^{\mathbb{P}}\bigl[\mathbb{E}^{\mathbb{P}}[Z_\infty \mid \mathcal{F}_t] \cdot Y_t \mid \mathcal{F}_s\bigr] = \mathbb{E}^{\mathbb{P}}[Z_t Y_t \mid \mathcal{F}_s] = Z_s Y_s, \end{align*} where the first equality uses that $Y_t$ is $\mathcal{F}_t$-measurable and pulls it out of the inner conditional expectation, the second uses the $\mathbb{P}$-martingale property of $Z$ ($\mathbb{E}^{\mathbb{P}}[Z_\infty \mid \mathcal{F}_t] = Z_t$), and the third uses the $\mathbb{P}$-martingale property of $ZY$. Substituting: \begin{align*} \mathbb{E}^{\mathbb{Q}}[Y_t \mid \mathcal{F}_s] = \frac{Z_s Y_s}{Z_s} = Y_s. \end{align*} This confirms $Y = \tilde{X}^{T_n}$ is a $\mathbb{Q}$-martingale for each $n$. Since $T_n \uparrow \infty$ $\mathbb{Q}$-a.s., $\tilde{X}$ is a continuous $\mathbb{Q}$-local martingale with reducing sequence $(T_n)$.[/step]

[guided]The final step transfers the martingale property across the change of measure. The tool is the **Bayes formula** (also called the abstract Bayes rule for conditional expectations): if $\mathbb{Q} \ll \mathbb{P}$ with Radon-Nikodym derivative $L = d\mathbb{Q}/d\mathbb{P}$, and $L_t = \mathbb{E}^{\mathbb{P}}[L \mid \mathcal{F}_t]$ is the density process, then for any bounded $\mathcal{F}_t$-measurable $\xi$, \begin{align*} \mathbb{E}^{\mathbb{Q}}[\xi \mid \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[L_t \xi \mid \mathcal{F}_s]}{L_s}. \end{align*} In our setting, $L = Z_\infty = \mathcal{E}(M)_\infty$ and $L_t = \mathbb{E}^{\mathbb{P}}[Z_\infty \mid \mathcal{F}_t] = Z_t$ (since $Z$ is a uniformly integrable $\mathbb{P}$-martingale). Applying with $\xi = Y_t$ (which is bounded by $n$): \begin{align*} \mathbb{E}^{\mathbb{Q}}[Y_t \mid \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[Z_t Y_t \mid \mathcal{F}_s]}{Z_s}. \end{align*} We showed in the previous step that $ZY$ is a $\mathbb{P}$-martingale, so $\mathbb{E}^{\mathbb{P}}[Z_t Y_t \mid \mathcal{F}_s] = Z_s Y_s$. Dividing by $Z_s > 0$ (which is legitimate since $Z_s = \mathcal{E}(M)_s = \exp(M_s - \frac{1}{2}\langle M \rangle_s) > 0$) gives \begin{align*} \mathbb{E}^{\mathbb{Q}}[Y_t \mid \mathcal{F}_s] = Y_s. \end{align*} So $Y = \tilde{X}^{T_n}$ is a $\mathbb{Q}$-martingale. Why does the entire argument hinge on the product $ZY$ being a $\mathbb{P}$-martingale? The Bayes formula tells us that a $\mathbb{Q}$-martingale property for $Y$ is equivalent to a $\mathbb{P}$-martingale property for $ZY$. The Itô product rule shows $d(ZY)$ has no drift (the finite-variation parts cancel), so $ZY$ is a local martingale; the uniform integrability of $Z$ combined with the boundedness of $Y$ upgrades it to a true martingale. This is the standard "density trick" in the theory of absolutely continuous changes of measure. Since $T_n \uparrow \infty$ $\mathbb{Q}$-a.s. and each $\tilde{X}^{T_n}$ is a $\mathbb{Q}$-martingale, the process $\tilde{X} = X - \langle X, M \rangle$ is a continuous $\mathbb{Q}$-local martingale, as claimed.[/guided]