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Suppose we are given a left continuous process $X=(X_t)_{t\ge 0}$ and define

$$Y^n_t=n\int^t_{t-\frac{1}{n}}\mathbf1_{\{|X_{s\vee 0}|\le n\}}X_{s\vee 0}ds$$

Why does it hold that $\lim_nY^n\to X$? It should follow from the left-continuity. Clearly this is important by the definition of the boundaries of the integral.

math
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1 Answers1

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After the substitution $u:=n(t-s)$ we obtain $$\large Y_t^n=\int_0^1\chi_{\{|X_{t-u/n\vee 0}|\leqslant n\}}X_{t-u/n\vee 0}\mathrm du,$$ and the role of left-continuity is clearer.

Davide Giraudo
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  • Thanks for your answer. By left continuity we have $\lim_n X_{t-\frac{u}{n}}=X_t$. But why can I interchange limit and integration? Also, this should just work for left continuous processes. However what exactly fails if we assume right continuity and define $Y_t^n:=\int_t^{t+\frac{1}{n}}\dots$? – math Feb 27 '14 at 15:20