Plug-in estimator of expected value

Let $g$ be the statistical functional defined by $g(\mu) = \int x \,d\mu$. Then the plug-in estimator is defined as $\hat{g}=g(L_n)$ where$$L_n(\omega)=\frac 1 n \sum_{i=1}^n \delta_{X_i(\omega)}$$ is theempirical distribution. Here $X_i$ are independent probabilitydistributions. We have

$$\hat{g}=\bar{X_n}=\frac 1 n \sum_{i=1}^n X_i$$

I don't understand why the last formula is true. I think it should be pretty simple, but plugging in the definitions I get the following:

$$\hat{g}=g(L_n)=$$$$\int \frac 1 n \sum_{i=1}^n \delta_{X_i} \, d\mu = $$$$\frac 1 n \sum_{i=1}^n \int \delta_{X_i} \, d\mu$$

But I just don't see why we have $\int \delta_{X_i} \, d\mu = X_i$