Where does the extra $1_{X=x_n}$ come in discrete conditional expectation?
$$\mathbb{E}[Y|X=x]=\frac{\mathbb{E}[Y1_{X=x}]}{\mathbb{P}(X=x)}$$
But then I see that when $\Omega$ is partitioned into union of subsets, then
$$\mathbb{E}[Y|\sigma(X)]=\sum_n \frac{\mathbb{E}{[Y1_{X=x_n}]}}{\mathbb{P}(X=x_n)}\mathbb{1}_{X=x_n}$$
So where does this last $\mathbb{1}_{X=x_n}$.
https://www.stat.berkeley.edu/users/pitman/s205f02/lecture15.pdf
If we write: $$f\left(x\right)=\mathbb{E}\left[Y\mid X=x\right]=\frac{\mathbb{E}Y\mathbf{1}_{X=x}}{P\left(X=x\right)}$$ then: $$f\left(X\right)=\mathbb{E}\left[Y\mid\sigma\left(X\right)\right]$$
For a fixed $\omega\in\Omega$ with $X\left(\omega\right)=x_{k}$ we get:
$$\begin{aligned}\left(\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}\right)\left(\omega\right) & =\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}\left(\omega\right)\\ & =\frac{\mathbb{E}Y\mathbf{1}_{X=x_{k}}}{P\left(X=x_{k}\right)}\\ & =\mathbb{E}\left[Y\mid X=x_{n}\right]\\ & =f\left(x_{n}\right)\\ & =f\left(X\left(\omega\right)\right) \end{aligned} $$
So the conclusion is that:
$$f\left(X\right)=\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}$$ or equivalently: $$\mathbb{E}\left[Y\mid\sigma\left(X\right)\right]=\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}$$