= Inclusion indicator
{title2=$I_i=\mathbf 1_{\{i\in S\}}$}
An <inclusion indicator> records whether a specified population unit appears in a sample $S$:
$$
I_i=\mathbf1_{\{i\in S\}}=
\begin{cases}1,&i\in S,\\0,&i\notin S.\end{cases}
$$
It is an <indicator random variable>, with <expected value> $\mathbb E I_i=\pi_i$ and <variance> $\pi_i(1-\pi_i)$. Its randomness comes from sample selection; the population value attached to unit $i$ may be treated as fixed in a sampling calculation.<Unequal Probability Sampling in Penn State STAT 506>[Lesson 3, inclusion probabilities and inverse-probability total estimation]{ref}
<Inclusion indicators> turn a sum over a changing sample into a sum over fixed population labels:
$$
\sum_{i\in S}y_i=\sum_{i=1}^NI_i y_i,\qquad
\mathbb E\!\left[\sum_{i\in S}y_i\right]=\sum_{i=1}^N\pi_i y_i.
$$
If every $\pi_i>0$, the same identity immediately proves
$$
\mathbb E\!\left[\sum_{i=1}^N\frac{I_i y_i}{\pi_i}\right]=\sum_{i=1}^Ny_i.
$$
The inverse-inclusion-probability total is unbiased because $\mathbb E(I_i/\pi_i)=1$, not because the selected units must look typical in every realization.
Pair products connect the notation to <joint inclusion probabilities>: $\mathbb E(I_iI_j)=\pi_{ij}$, so $\operatorname{Cov}(I_i,I_j)=\pi_{ij}-\pi_i\pi_j$. For a fixed <sample size> $n$, $\sum_iI_i=n$ in every realization. Consequently,
$$
0=\operatorname{Var}\!\left(\sum_iI_i\right)
=\sum_i\operatorname{Var}(I_i)+2\sum_{i<j}\operatorname{Cov}(I_i,I_j).
$$
In <simple random sampling> with $N=4,n=2$, the four individual variances sum to $1$, while the six distinct pair covariances are each $-1/12$. Their doubled sum is $-1$, giving total variance zero. Treating the <inclusion indicators> as independent would incorrectly predict a random sample size. The notation makes that error easy to detect.
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