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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 $\int_0^1 f(x_i) di = \int^1_0 f(x_{ss}) di = f(x_{ss}) \int^1_0 di = f(x_{ss})$

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  • $\begingroup$ 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. $\endgroup$ Jul 17, 2023 at 21:02
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    $\begingroup$ $F$ is any arbitrary function. I have replaced $F$ with $f$ in the question. $\endgroup$
    – user43994
    Jul 18, 2023 at 9:56

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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 $$\int_0^1f(x_i)di=\int_0^1f(x(i))di=\int_0^1f(x_{ss})di=f(x_{ss})\int_0^1di=f(x_{ss})$$

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