# Can we get simultaneous convergence of the integral using simple functions?

Let $$Y$$ be an integrable nonnegative random random variable on some probability space $$X$$, and Let $$F:[0,\infty) \to \mathbb R$$ be a continuous function.

Suppose that $$F(Y) \in L^1(X)$$. (I am fine with assuming $$Y$$ is bounded as well).

Do there always exist simple functions $$Y_n \ge 0$$ on $$X$$ such that

$$E(Y)=\lim_{n \to \infty} E(Y_n), E(F(Y))=\lim_{n \to \infty} E(F(Y_n))$$ both hold simultaneously?

Taking $$Y_n$$to be increasing, we get the first equality, due to the monotone convergence theorem. But if $$F$$ is not increasing, then $$F(Y_n)$$ won't necessarily be increasing.

• Are both $Y$ and $F(Y)$ integrable? – zhoraster Jun 3 at 17:05
• Yes, I have updated the question to mention this explicitly. – Asaf Shachar Jun 3 at 19:06

Under the assumption $$Y$$ is bounded: There exists a positive real $$M$$ such that $$0\le Y \le M$$ on $$X.$$ Let $$Y_n$$ be the usual simple functions approximating $$Y.$$ We then have $$0\le Y_n\le M$$ for all $$X.$$ Furthermore, $$Y_n\to Y$$ uniformly on $$X;$$ this you will recall holds because $$Y$$ is bounded. Now $$F$$ is uniformly continuus on $$[0,M],$$ and from this it follows that $$F\circ Y_n\to F\circ Y$$ uniformly on $$X.$$ The measure on $$X$$ is finite and the result follows.

You don't need $$Y$$ to be bounded. First, we can assume WLOG that $$F$$ is non-negative and $$F(0)=0$$. Let $$(\varepsilon_n)$$ be a sequence decreasing to $$0$$ and $$A_n := \{Y \le n \}$$. Since $$F$$ is uniformly continuous on $$[0,n]$$ for every $$n$$, there exists $$\delta_n$$ such that if $$|x-y| < \delta_n$$ and $$0 \le x,y \le n$$ we have $$|F(x)-F(y)| < \varepsilon_n$$.

Take $$Y_n \ge 0$$ to be a simple function such that $$Y_n \le Y$$, $$Y_n = 0$$ on $$A_n^c$$, and $$|Y-Y_n| < \delta_n$$ on $$A_n$$. Then we have

$$0 \le \mathbb{E}[Y-Y_n] = \mathbb{E}[(Y - Y_n)1_{A_n}] + \mathbb{E}[Y1_{A_n^c}] < \delta_n + \mathbb{E}[Y1_{A_n^c}].$$

The first term goes to $$0$$ by our choice of $$\delta_n$$, and the second term goes to $$0$$ by the monotone convergence theorem, so $$(Y_n) \rightarrow Y$$ in $$L^1$$. Similarly, we compute

$$\mathbb{E}[|F(Y)-F(Y_n)|] = \mathbb{E}[|F(Y)-F(Y_n)|1_{A_n}] + \mathbb{E}[|F(Y)-F(0)|1_{A_n^c}] < \varepsilon_n + \mathbb{E}[|F(Y)|1_{A_n^c}]$$

and again the first term goes to $$0$$ by our choice of $$\varepsilon_n$$ and the second term goes to $$0$$ by the monotone convergence theorem.