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If $ M$ is a continuous local martingale, then it exists a sequence of partitions of $[0,\infty)$ with $|\Pi_n| \to 0 $ ($|\Pi|$ denotes the mesh side) such that

$$ P(\lim_{n\to \infty} Q^{\Pi_n}_t = \langle M \rangle_t \mbox{ simulataneously for all t $\ge$ 0 }) = 1$$

Where, $Q^{\Pi_n}_t:=\sum_{t_i \in \Pi_n} (M_{t_{i+1}\wedge t}-M_{t_i\wedge t})^2$ for $t\ge 0$.

I have two question about the proof:

  1. If we know, $\lim_{|\Pi|\to 0} Q_t^\Pi = \langle M \rangle_t$ in probability, we deduced that using a diagonal procedure we have a sequence $(\Pi_n)_n$ with

$$P(\lim_{n\to \infty} Q^{\Pi_n}_q = \langle M \rangle_q \mbox{ simulataneously for all rational q $\ge$ 0 }) = 1$$

Diagonal argument is clear. But here we use also the fact, that for a converging sequence in probability, there exists a subsequence which converges a.s., right?

  1. Why is this enough to show? They say, because $\langle M \rangle$ is continuous. Could someone explain this in more detail?



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up vote 2 down vote accepted

Hint: Consider some nondecreasing functions $g_n$ and a continuous function $g$ defined on $[0,+\infty)$ such that $g_n(x)\to g(x)$ when $n\to\infty$, for every $x$ in some dense subset of $[0,+\infty)$. Then $g_n(x)\to g(x)$ when $n\to\infty$, for every $x$ in $[0,+\infty)$.

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@ did: thanks for your answer. I thought that, but was not quite sure! Are my thoughts in the first question correct? – math May 30 '12 at 18:20

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