Given XtX_t Is increasing And deterministic function f(t)f(t) , do we have

E(∫t0f(u)dXu)=∫t0f(u)dE(Xu)?E\left(\int_{0}^{t}f(u)\mathrm dX_u\right) =\int_{0}^{t}f(u)\mathrm dE(X_u)?

I think it goes well as it works for each approximating sum.

If not, any extra condition to make it holds?

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