# What rules apply integrals over a continuum of agents in a symmetric equilibrium?

Suppose I have a model with a continuum of agents denoted by $$i \in [0;1]$$. In a symmetric equilibrium, i.e. in a equilibrium where $$x_i = x_j$$ for all $$x,i$$, would I be able to conclude that: and some expression which I know holds: $$\int^1_0 f(x_i) di = f \left( \int^1_0 x_i di \right)$$ The reasoning being that $$\int^1_0 f(x_i) di$$ can heuristically be thought of as a normalized sum over agents, i.e. as: $$\sum_i^n \frac{1}{n} f(x_i)$$ Which in a symmetric equilibrium equals $$\frac{n}{n} f(x_i) = f(x_i) = f\left(\sum_i^n \frac{1}{n} x_i\right)$$ Which can again heuristically be thought of as $$f \left( \int^1_0 x_i di \right)$$.

Are my heuristics wrong? Also if someone has a more rigorous approach to understanding these types of integrals please tell me. I have been unable to find a better answer than "think of the integral as a sum" while searching this website and others.

EDIT: I believe this equality does indeed hold. We can think of $$x$$ as a function of $$i$$. In a symmetric steady-state the function $$x(i)$$ is simply the constant function which returns the constant $$x_{ss}$$, thus $$\int_0^1 f(x_i) di = \int^1_0 f(x_{ss}) di = f(x_{ss}) \int^1_0 di = f(x_{ss})$$

• What is $F$? This formulas seem related to some theorem of passage of the limit or of a derivative under the integral sign. In general, these theorems have to do with measure theory, which is used in probability, and the Lebesgue's integral. Jul 17, 2023 at 21:02
• $F$ is any arbitrary function. I have replaced $F$ with $f$ in the question. Jul 18, 2023 at 9:56

As I've written in the edit I believe this equality does indeed hold. We can think of x as a function of i. In a symmetric steady-state the function $$x(i)$$ is simply the constant function which returns the constant $$x_{ss}$$, thus $$\int_0^1f(x_i)di=\int_0^1f(x(i))di=\int_0^1f(x_{ss})di=f(x_{ss})\int_0^1di=f(x_{ss})$$