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Let $Z$ be a random variable with finite mean and consider the function $F: R \rightarrow R$ defined by $ F(x) = E[\max(x+Z,0)]$ How to prove that $F(x)$ is a continuous function of $x$?

This is not homework. A textbook I am reading states that $F(x)$ is continuous without providing any explanation, and I can't seem to fill the gap. I would be happy with a hint.

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If $\lvert x-y\rvert\leq\delta$, then $\lvert (x+Z)-(y+Z)\rvert\leq\delta$, and also $\lvert \mathrm{max}(x+Z,0)-\mathrm{max}(y+Z,0)\rvert\leq\delta,$ right? (Notice that $\mathrm{max}(\bullet+Z,0)$ is 1-Lipschitz.)

Hence we also have $\lvert F(x)-F(y)\rvert\leq\delta$.

Remark: In order for $\mathrm{max}(x+Z,0)$ to be well-defined, we should assume that $Z$ is bounded, I guess. (According to the comments this is not neccessary.)

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    @robinson: It suffices to assume $Z$ has finite mean. Indeed, by conditioning on $Z$, we have $E[\max(x+Z,0)] = \int_{( - x,\infty )} {(x + s)F_Z (ds)}$, where $F_Z$ is the distribution of $Z$.2010-11-30