The proof I am concerned is 2.2.5 pg 37 Kochman Stable Homotopy.
Let $R$ be a commutative ring. Let $S^n \rightarrow E \xrightarrow{p} B$ be Serre fibration. $B$ a CW complex, $B$ simpliy connected. Define a filtration, $E^n:=p^{-1}(B^n)$ of $E$. Then $$\varprojlim H^*(E^n;R) = H^*(E;R)$$
So my attempt is to show $H^i(E^{n+1})\rightarrow H^i(E^n)$ is an isomorphism for all large $n$.
1. $\pi_i(B,B^n)=0$ for $n$ sufficiently large by cellular approximation.
If $(B,B^n)$ is $k-1$-connected, then so is $(E,E^n)$ by lifting property
By Hurewicz theorem $\pi_k(E,E^n)= H_k(E,E^n)$, which is zero. So by LES, $H_k(E^n;\Bbb Z) \rightarrow H_k(E; \Bbb Z)$ is isomorphism.
By UCT, there is exact sequence $$ Tor(H_{i-1}(X;\Bbb Z);R) \rightarrow H^i(X; R) \rightarrow Hom(H_i(X;\Bbb Z);R)$$ which is functorial in its argument. Hence, this induces surjection on cohomology $H^k(E;R) \rightarrow H^k(E^n;R)$
Comments: So none of this proof requires the ring $R$ to be integral?
Mittag Leffler condition: if for each $t$, $j^*_n:H^t(X^{n+1}) \rightarrow H^t(X^n)$ satisfies Mittag Leffler, then
$$H^*(X) = \varprojlim H^*(X^n)$$
This is page 123 in Kochman's stable homotopy theory and adams spectral sequence.