Let $X_1,X_2,.....,X_n$ deonte the i.i.d random variables for a random sample of size $n$.Then the mean of sample is a random variable defined by $$\bar{X}=\frac{X_1+X_2+.....+X_n}{n}$$
Let $f(x)$ be the pdf of some given population from which we draw a sample of size $n$,We have that "The mean of the sampling distribution of means denote by $E[\bar{X}]=\mu$" ,where $\mu$ is the mean of the population. I want to know the proof of this, below is what I tried.
$E[\bar{X}]=E[\frac{X_1+X_2+.....+X_n}{n}]=\frac{1}{n}E[X_1+X_2+....+X_n]=\frac{1}{n}\{E[X_1]+E[X_2]+....+E[X_n]\}$
Now I think the reason that each $E[X_i]=\mu$ is that since all $X_i's$ are i.i.d, But I want to know why this is, like some mathematical proof for this that all $E[X_i]=\mu$.