About block of category $\mathcal{O}$ In the book "Representation of Semisimple Lie Algebra in the BGG Category $\mathcal{O}$".
Exercise 1.13. Suppose $\lambda\not\in\Lambda$, so the linkage class $W\cdot\lambda$ is the disjoint union of its nonempty intersections with various cosets of $\Lambda_r$ in $\mathbb{h}^*$. Prove that each $M\in\mathcal{O}_{\chi_\lambda}$ has a corresponding direct sum decomposition $M=\oplus_i M_i$, in which all weights of $M_i$ lies in a single coset.
My attempt: 
$M=\bigoplus_{\nu\in\mathfrak{h}^*} M_\nu
=\bigoplus_{[\nu]\in\mathfrak{h}^*/\Lambda_r} M^{[\nu]}$.
Since $\mathfrak{g}_\alpha\cdot M_\mu\subseteq M_{\mu+\alpha}$ for all $\alpha\in \Phi$, we get $U(\mathfrak{n})\cdot M^{[\nu]}\subseteq M^{[\nu]}$, $U(\mathfrak{h})\cdot M^{[\nu]}\subseteq M^{[\nu]}$ and $U(\mathfrak{n}^-)\cdot M^{[\nu]}\subseteq M^{[\nu]}$. 
Since $U(\mathfrak{g})=U(\mathfrak{n}^-)U(\mathfrak{h})U(\mathfrak{n})$, we get 
$U(\mathfrak{g})\cdot M^{[\nu]}\subseteq M^{[\nu]}$. Hence $M^{[\nu]}$ is a $U(\mathfrak{g})$-submodule of $M$.
Since $M\in\mathcal{O}$, $M$ is finitely generated as a $U(\mathfrak{g})$-module. Therefore, $M=\bigoplus_{i=1}^n M^{[\nu_i]}$.
Now, let $W\cdot\lambda=\{\eta_1,\cdots,\eta_k\}$. 
Consider $\{\eta_{i_1},\cdots, \eta_{i_r}\}\subseteq \{\eta_1,\cdots,\eta_k\}$ such that $[\eta_{i_1}],\cdots, [\eta_{i_r}]$ are distinct and $\{[\eta_{i_1}],\cdots, [\eta_{i_r}]\}=\{[\eta_1],\cdots,[\eta_k]\}$. It is clearly that $W\cdot\lambda\cap[\eta]\neq \emptyset\implies [\eta]\in \{
[\eta_{i_1}],\cdots, [\eta_{i_r}]\}$.
Then $W\cdot \lambda
=\bigcup_{\eta\in\mathfrak{h}^*} W\cdot\lambda \cap[\eta]
=\bigsqcup_{j=1}^{r}W\cdot\lambda \cap[\eta_{i_j}]
=\bigsqcup_{j=1}^{r}W\cdot \eta_{i_j} \cap[\eta_{i_j}]
=\bigsqcup_{j=1}^{r}W_{[\eta_{i_j}]}\cdot \eta_{i_j}$
I would like to know whether the corresponding direct sum decomposition means $M=\bigoplus_{j=1}^r M^{[\eta_{i_j}]}$.
If so, how to prove it? Also, am I on the right track?
 A: I think you're overcomplicating things. You have showed that: (a) As a vector space we have the following decomposition:
$$
M = \bigoplus_{[\nu]\in\mathfrak{h}^*/\Lambda_r} M^{[\nu]}.
$$
(b) Furthermore, each $M^{[\nu]}$ is a submodule of $M$.
From this it follows directly that the direct sum above is a decomposition of modules.
Edit to clarify:
Note that just before Exercise 1.13, Humphreys writes "The following easy exercise gives a refinement of the linkage classes in the nonintegral case..." (emphasis mine). Chapter 1.1 has just introduced Category $\mathcal{O}$, and Exercise 1.1(b) shows that $\mathcal{O}$ decomposes as 
$$\mathcal{O}=\bigoplus_{[\nu]∈\mathfrak{h}^∗/\Lambda_r}\mathcal{O}^{[\nu]}.$$
Chapter 1.12 shows that $\mathcal{O}$ also decomposes by central character as
$$\mathcal{O}=\bigoplus_{\chi}\mathcal{O}_\chi.$$
Chapter 1.13 introduces the notion of a block of a category, and shows that if $\lambda$ is integral then $\mathcal{O}_{\chi_\lambda}$ is a block. Exercise 1.13 is about what happens when $\lambda$ is not integral, and is a straightforward application of the result of Exercise 1.1(b). In this case $[w\cdot\lambda]$ is not a single coset of $\Lambda_r$ in $\mathfrak{h}^*$, but several, say $[w\cdot\lambda] = \{[\nu_1], \dots, [\nu_k]\}$. Exercise 1.1(b) now tells us that
$$\mathcal{O}_{\chi_\lambda}=\bigoplus_{i=1}^k\mathcal{O}_{\chi\lambda}^{[\nu_i]},$$
so when $\lambda$ is not integral then $\mathcal{O}_{\chi_\lambda}$ splits into several blocks, each corresponding to one of the cosets in $[w\cdot\lambda]$.
A: In Humphreys' book, I quote "A formal argument (which we leave to the reader) shows that
M decomposes uniquely as a direct sum of submodules, each belonging
to a single block. In particular, each indecomposable module belongs to a
single block. (See for example Jantzen [153, II.7.1]; his finite dimensionality
assumption can be replaced here by the chain conditions on modules.)"
Remark: Now if $M$ is arbitrary, we say it belongs to a block if all its composition
factors do.
The blocks of $\mathcal{O}$ are precisely the subcategories consisting of modules whose composition factors all have highest weights linked by $W_{[\lambda]}$ to an
antidominant weight $\lambda$. We denote the block associated to an antidominant
weight by $\mathcal{O}_\lambda$.
By mimicking the proof in Jantzen [153, II.7.1]:
Let us denote the set of blocks by $\mathcal{B}$. For any $M\in \mathcal{O}_{\chi_\lambda}$, let $M_b$ be the sum of all submodules $M'$ of $M$ such that all composition factors of $M'$ belong to $b$. Then $M_b$ is the largest submodule with this property. By the remark above, we get $M_b\in b$.
Claim $M=\bigoplus_{b\in\mathcal{B}}M_b$.
The sum is clearly direct. It suffices to show $N=\sum_{b\in\mathcal{B}}M_b$ is all of $M$.
Suppose $N\neq M$. Then $\{0\}\neq M/N\in \mathcal{O}$, by corollary 1.2. in Humphreys' book, there is a chain $\{0\}\subseteq N'/N\subseteq M/N$ such that $N'/N=L$ is simple.
Equivalently, there is a submodule $N'\subseteq M$ such that $N\subset N'$ such that $N'/N$ is simple. There is a block $b$ with $L\in b$. 
Set $N''=\bigoplus_{b'\neq b}M_{b'}\subseteq N\subset N'$. Then $\{0\}\subseteq N/N''\subseteq N'/N''$. 
All composition factors of $N'/N''$ belong to $b$ since they are either $L\cong (N'/N'')/(N/N'')$ or composition factors of $N/N''\cong M_b$. 
By the definition of blocks, the exact sequence $0\to N''\to N'\to N'/N''\to 0$ splits. So there is a submodule $E$ of $N'$ with $N'=N''\oplus E$. We have $E\cong N'/N''$, hence $E\subseteq M_b$. It follows that $N'\subseteq \bigoplus_{b'}M_{b'}=N$, a contradiction. Therefore, the claim follows.
Now claim all weights of $M_b\in b$ lies in a single coset.
If $M_b$ is indecomposable, by exercise 1.1 (b) in Humphreys' book, we are done.
If $M_b$ is decomposable, then $M_b=M_b^1\oplus\cdots\oplus M_b^k$, where $M_b^i$ is indecomposable. Note that $M_b^i\in b$ since $M_b\in b$.
Let $b=\mathcal{O}_\lambda$. By corollary 1.2., there is a composition series for $M_b^i$: $\{0\}\subseteq M_1^i\subseteq \cdots\subseteq M_b^i$. Then $M_1^i$ have a highest weight in the form $w_i\cdot\lambda$ with $w_i\in W_{[\lambda]}$. This implies $M_b^i$ contains a weight in the form $w_i\cdot\lambda$ with $w_i\in W_{[\lambda]}$. By exercise 1.1 (b), all weights of $M_b^i$ lie in $[w_i\cdot\lambda]$ for each $i$. Since $w_i\in W_{[\lambda]}$, we get $w_i\lambda-w_j\lambda\in\Lambda_r$ and then $[w_i\cdot\lambda]=[w_j\cdot\lambda]$ for any $i,j$. Hence all weight of $M_b$ lie in $[w\cdot\lambda]$ where $w\in W_{[\lambda]}$.
