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I'm a little uneasy about one step in the uniqueness proof for Doob's Decomposition Theorem. Let $(X_n)_{n \geq 0}$ be a submartingale, $(M_n)_{n \geq 0}$ a martingale, and $(A_n)_{n \geq 0}$ be an increasing process such that $A_{n+1}$ is $\mathcal{F}_n$-measurable. If we have 2 decompositions \begin{align*} X_n &= X_0 + M_n + A_n\\ X_n &= X_0 + L_n + C_n, \end{align*} Then $M_n - L_n = A_n - C_n$ (a.s.?).

Here is the part I don't get: My book argues that since $A_n - C_n$ is $\mathcal{F}_{n-1}$-measurable, $M_n - L_n$ is also $\mathcal{F}_{n-1}$-measurable. Why is this true? I feel like in general, $X=Y$ a.s. and $Y$ is $\mathcal{F}$-measurable does not imply that $X$ is $\mathcal{F}$-measurable, like the example here:Almost sure convergence for measurability

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The equality $$M_n-L_n = A_n-C_n$$ does not only hold almost surely, but (pointwise) for each $\omega$, i.e. $$M_n(\omega) - L_n(\omega) = A_n(\omega)-C_n(\omega) \qquad \text{for all} \, \, \omega \in \Omega.$$ Therefore the left-hand side has exactly the same measurability properties as the right-hand side.


An alternative way to prove uniqueness goes as follows: From the given decomposition it is not difficult to see that $$\mathbb{E}(X_j-X_{j-1} \mid \mathcal{F}_{j-1}) = A_j-A_{j-1}.$$ Summing over $j=1,\ldots,n$ gives $$A_n = \sum_{j=1}^n \mathbb{E}(X_j-X_{j-1} \mid \mathcal{F}_{j-1}).$$ Consequently, $$M_n = X_n-X_0-A_n$$ also holds true. This shows that $(A_n)_n$ and $(M_n)_n$ are unique (we even have an explicit formula for both of them).

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  • $\begingroup$ Thank you for the thorough explanation! If I could bother you with one last quick question, do we know that $M_n - L_n = A_n - C_n$ for all $\omega$ because $X_n - X_n = 0$ for all $\omega$? $\endgroup$
    – user207886
    Feb 6, 2015 at 14:56
  • $\begingroup$ @user207886 Yes, that's correct. $\endgroup$
    – saz
    Feb 6, 2015 at 14:59

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