The context is a finitely generated modules of a finite dimensional $\mathbb{K}$-algebra $A$, where $\mathbb{K}$ is an arbitrary field.
Now, if we consider $A$ as a left $A$-module as usual, this module $A$ has many composition series which, according to the Jordan-Holder theorem, have the simple quotients in common, meaning that each simple module appears the same number of times in any series (and since any simple module has to be a quotient of $A$ these are also all the simple modules).
We also have the notion of the radical series of $A$ $$A \supseteq \operatorname{rad}(A) \supseteq \operatorname{rad}(A)^2 \supseteq \dots \supseteq \operatorname{rad}(A)^l \supseteq 0 $$ in which, due to the basic properties of the radical and the fact that $\operatorname{rad}^n(A) = \operatorname{rad}(\operatorname{rad}^{n-1}(A))$ we have that each quotient is a semisimple $A$-module. Decomposing these quotients in direct sums of simple modules, we get a list of simple modules.
My question is if all isomorphism types of simple $A$-modules also appear in this list, and if their multiplicity here is the same as the one they would have in any composition series. I think the answer is yes to both questions, but I am not able to show this.
All isotypes of simple left $A$-modules already appear in $A/rad(A)$. This is because the two rings have the "same set" of simple modules, and since $A$ is Artinian, each one is a direct summand of $A/rad(A)$.
Since you can take any of those semisimple radicals and express it as a composition series, you can chain them all together to get a composition series for $A$, so yes, the multiplicities of simple modules appearing throughout are governed by the uniqueness of the composition series for $A$.