Why is $F_{n,0}=H_n(K)$ for an arbitrary filtered complex?

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Let $... \subset K_{-1}=0 \subset K_0\subset ...K_n \subset...$ be an arbitrary filtered chain complex with $colim_n K_n:=K$.

Let $F_{p,p+q}=im(H_{p+q}(K_p) \to H_{p+q}(K))$

Mosher and Tangora decided to write down on page 67 that $F_{n,0}:H_n(K_n)=H_n(K)$.

First this is not even true for the filtered chain complex $K_n= C_*(X_n)$ with $X_n$ the $n-$ skeleton of a CW complex $X$; the correct statement in this case is that the induced map $H_n(K_{n+1})=H_n(K)$ is an iso. Moreover, this is not true in general because I can construct a stupid filtered complex s.t. $K_i=K_{i+1}=K_{i+....}$ however many times I would like.

What is the correct statement? I am familiar with spectral sequences in that I have done many computations with them. One of them is on my stackexchange.