Tensor product of irreducible representations of semisimple Lie algebras Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be semisimple Lie algebras over $\mathbb{C} $. Let $V_1$ and $V_2$ be simple modules  over $\mathfrak{g}_1$ and $\mathfrak{g}_2$, respectively. They can be infinite dimensional. 
$V_1 \otimes V_2$ is a $\mathfrak{g}_1 \oplus \mathfrak{g}_2$-module with the action given by $$ (x_1, x_2)(v_1 \otimes v_2)=(x_1 v_1)\otimes v_2 + v_1 \otimes (x_2 v_2). $$
Is it true that $V_1 \otimes V_2$ is a simple $\mathfrak{g}_1 \oplus \mathfrak{g}_2$-module?  If yes, how do I prove it? 

This is not such a trivial problem. The solution I know uses the Jacobson density theorem:
Let $M$ be a simple $R$-module, $D = \mathrm{End}_R(M)$. If $\varphi \in \mathrm{End}_D(M)$, $x_1,\ldots, x_n \in M$, then there exists $r \in R$ such that $rx_i = \varphi(x_i)$ for all $i$. I will show how this reduces to the case of principal tensors.
Let $\mathcal U_1$ and $\mathcal U_2$ be the enveloping algebras of $\mathfrak g_1$ and $\mathfrak g_2$. By Dixmier's Lemma, even if $V_1$ and $V_2$ are infinite-dimensional,
$$ \mathrm{End}_{U_i}(V_i) = \mathbb C$$
for $i = 1,2.$ (This is a generalization of the usual finite-dimensional Schur lemma to modules of countable dimension over $\mathbb C$).
Now suppose that $$v = \sum_{i=1}^n v^1_i \otimes v^2_i \in V_1 \otimes V_2$$ is nonzero. We want to show $\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2) v = V_1 \otimes V_2$.
Without loss of generality, we may choose $\{v^1_i\}$ to be linearly independent over $\mathbb C$ and $v^2_i \neq 0$ for all $i$. By Jacobson density, there exists $u \in \mathcal U_1$ such that 
$$ uv^1_i = \begin{cases} v^1_1 & i= 1 \\ 0 & i > 1 .\end{cases}$$
Hence for $u\otimes 1 \in \mathcal U_1 \otimes \mathcal U_2 =\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2)$,
$$(u \otimes 1)v = v^1_1 \otimes v^2_1,$$ 
which reduces to the case of when $v$ is a principal tensor.
