Unitary representation of $SO(3)$ 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
 A: We have
$$
\int_{\mathbb R^3} [\Pi_1(R) f](x)\overline{[\Pi_1(R) g]}(x) dx = \int_{\mathbb R^3} f(R^{-1} x) \bar g(R^{-1}x) dx.
$$
Now make the substitution $u = R^{-1} x$.  Since $R \in SO(3)$ the Jacobian of this transformation is 1.  So the above is $\int_{\mathbb R^3} f(u) \bar g(u) du$.  This shows that each $\Pi_1(R)$ is a unitary operator since it preserves the $L^2$ inner product.
A: 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?)
A: To show that $\Pi_1(R)$ is unitary you have to prove:


*

*$\langle \Pi_1(R) f, \Pi_1(R) g \rangle = \langle f, g \rangle$ for each $f, g\in L^2(\mathbb R^3)$

*$\Pi_1(R)$ is surjective.


For each $f, g\in L^2(R^3)$ we have
$$
\langle \Pi_1(R)f, \Pi_1(R)g \rangle := \int_{R^3} f(R^{-1}x)\overline{g(R^{-1}x)} dx = \int_{R^3} f(x)\overline{g(x)} dx =: \langle f, g \rangle
$$
Writing the previous equality we used the rotationally invariance of Lebesgue measure.
For each $f\in L^2(R^3)$, let $\tilde f$ be the function $x \to f(R x)$, we have 
$$
(\Pi_1(R) \tilde f)(x) = \tilde f(R^{-1}x)= f(R R^{-1}x) = f(x)
$$
So both requirements are satisfied. 
