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Assume $\displaystyle\lim_{t\to0}X_t=\gamma\hspace{3pt}a.s.$ where $X_t\geq 0$. I would like to show that $\displaystyle\lim_{t\to0}E[X_t]=E[\lim_{t\to0}X_t]=\gamma$, i.e. that it's possible to interchange the order of limit and expectation.

It would be sufficient to show $\displaystyle\lim_{t\to0}E[|X_t-\gamma|]=0$. I can't find a way to use the dominated/monotone convergence theorems, so I am wondering whether the following argument is valid:

$\displaystyle\lim_{t\to0}E[|X_t-\gamma|] =\lim_{t\to0}E[|X_t-\gamma|{\bf 1}\{|X_t-\gamma|<M\}] +\lim_{t\to0}E[|X_t-\gamma|{\bf 1}\{|X_t-\gamma|>M\}]$

For the first term we have by using the dominated convergence theorem and the fact that $X_t\to\gamma$ with probability one: \begin{equation} \begin{aligned} & \displaystyle\lim_{t\to0}E[|X_t-\gamma|{\bf 1}\{|X_t-\gamma|<M\}]\\ & =E[\lim_{t\to0}|X_t-\gamma|{\bf 1}\{|X_t-\gamma|<M\}]\\ & =0 \end{aligned} \end{equation}

For the second term we have by the inverse of Fatou's lemma and the fact that $X_t\to\gamma$ with probability one: \begin{equation} \begin{aligned} & \lim_{t\to0}E[|X_t-\gamma|{\bf 1}\{|X_t-\gamma|>M\}]\\ & \leq\limsup_{t\to0}E[|X_t-\gamma|{\bf 1}\{|X_t-\gamma|>M\}]\\ & \leq E[\limsup_{t\to0}|X_t-\gamma|{\bf 1}\{|X_t-\gamma|>M\}]\\ & \leq E[\lim_{t\to0}|X_t-\gamma|{\bf 1}\{|X_t-\gamma|>M\}]\\ & =E[0]\\ & =0 \end{aligned} \end{equation}

Combining we have $\displaystyle\lim_{t\to0}E[|X_t-\gamma|]=0$

Is this a valid solution or am I cheating somewhere?

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1 Answer 1

up vote 1 down vote accepted

The limsup version of Fatou's lemma only holds if there exists an integrable function $g$, such that $f_n < g$ for all n. In that case, your proof is valid; if not, there are counter-examples (e.g. $f_n = \gamma + n * 1_{[0,1/n]}$ )

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