I am reading Humphreys's book GTM9 on Lie algebra representations. The question above seems quite natural but the book doesn't give any clues on this topic, as far as I can see. Let's adopt the settings in the book: algebracally closed field with characteristic 0, finitely dimensional semisimple Lie algebra, and finitely dimensional represetations. Here by unique we mean the decomposition is unique up to order.
Is the decomposition of finitely dimensional representation of a semisimple Lie algebra into irreducible ones unique?
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The meaning of the question is not unique.
Literally understood I'd say that the answer is no, as soon as one irreducible occurs with multiplicity $\ge 2$. For instance, if $V$ is irreducible, then $V\oplus V$ decomposes both as the given sum, and as the sum of $V\times\{0\}$ plus the submodule $\{(tv,v):v\in V\}$ for each given scalar $t$, yielding, up to order as many decompositions as scalars.
When there is no multiplicity $\ge 2$, the decomposition is indeed unique up to order (i.e., the (unordered) set of summands is unique: these are precisely the irreducible subrepresentations).
What is true however is a weaker form of uniqueness occurring in general, namely up to isomorphism. Namely for every finite-dimensional representation $V$ with two irreducible decompositions $V=\bigoplus_{i\in I}V_i=\bigoplus_{j\in J}W_j$, there is a bijection $\sigma:I\to J$ such that $V_i$ and $W_{\sigma(i)}$ are isomorphic representations for each $i$.
Yes, it is unique up to order. You can see this by observing that the number of times an irreducible representation $S$ appears as a summand in a representation $V$ is exactly $\operatorname{dim} \operatorname{Hom}_{\mathfrak{g}}(S,V)$ (by Schur lemma). And the latter expression is independent of the choice of decomposition of $V$ into simples.
EDIT: As Jason DeVito suggested in the comments, uniqueness does not mean that you can find unique simple submodules of $V$ that make the decomposition of $V$ into simples. But the simples are unique up to isomorphism, of course. (And in fact unique in the strong sense if and only if all the multiplicities are $1$.)