Can we extend convex combination to integral?

Question

The definition of the convex combination is given as: $$x=\lambda x_1 + (1-\lambda) x_2,\forall \lambda \in [0,1]$$ It can be obviously extend to multi-terms form as: $$x=\sum_{i=1}^N \lambda_i x_i,\sum_{i=1}^N \lambda_i=1,\lambda_i \ge 0$$ I wonder can we extend it to a continous form as: $$x=\int xp(x)dx,\int p(x)=1,p(x)\ge 0$$ If we can get the contious version. We can then prove the probability inequality: $$f(E_p[x])\le E_p[f(x)]$$

Answer 1

Yes, under appropriate conditions this works. More generally, if $\mu$ is a probability measure on $X$ and $f$ is a $\mu$-integrable function, we can approximate $\int_X f \; d\mu$ by convex combinations $\sum_{j} f(x_j) p_j$ with $p_j \ge 0$ and $\sum_j p_j = 1$.

Answer 2

This is called Jensen's inequality.