TITLE: Unitary representation of $SO(3)$
QUESTION [5 upvotes]: Definition: $\mathcal{H}$ be a Hilbert space and $U(\mathcal{H})$ denote the unitary operators on it, If Unitary representation of a matrix lie group $G$ is just a homomorphism $\Pi:G\rightarrow U(\mathcal{H})$ with the following continuity condition: $A_n\rightarrow A\Rightarrow \Pi(A_n)v\rightarrow\Pi(A)v$
Now could any one help me what is going on here in detail so that I can understand,  
"let $\mathcal{H}=L^2(\mathbb{R}^3,dx)$ the space of all square integrable functions on $\mathbb{R}^3$, for each $R\in SO(3)$ we define an operator $[\Pi_1(R)f](x)=f(R^{-1}x)$, since Lebesgue measure is rotationally invariant, $\Pi_1(R)$ is a unitary operator for each $R\in SO(3)(why?)$ , and it is easy to show $R\rightarrow \Pi_1(R)$ is unitary representation."
Thank you

REPLY [3 votes]: Consider SO(3) generators:
$$
[X_i,X_j]=i \epsilon^{ijk} X_k\\
X_1=i {\begin{pmatrix}  0& 0&0\\ 0& 0&-1\\0&1&0  \end{pmatrix}}\\
X_2=i {\begin{pmatrix} 0& 0&1\\ 0& 0&0\\-1&0&0  \end{pmatrix}}\\
X_3=i {\begin{pmatrix} 0& -1&0\\ 1& 0&0\\0&0&0  \end{pmatrix}}
$$
Note $X_j^\dagger=X_j$.
Write the SO(3) Lie group elements as 
$$
g=e^{i \alpha_j X_j}
$$
where $g^{-1}=g^\dagger$. And $g^{-1}g=g^\dagger g=1$.
So $g=e^{i \alpha_j X_j}$ is the unitary Rep of SO(3). (Just make sure my statement here is correct?)