Problem Let $X,Y:(\mathbb{R},\mathscr{B}(\mathbb{R}),P)\mapsto (\mathbb{R},\mathscr{B}(\mathbb{R})),$ and $P = f\cdot m$ where $f: \mathbb{R} \mapsto \mathbb{R}, f(x)=1_{(0,1)}(x)$ is the uniform density. X and Y are two random variables defined by $$X(x)=2x, Y(x)=2x\cdot 1_{(0,1)}(x)$$
Is it true $X = Y \text{P-a.s}?$
Attempt: From a previous question it's known that $X = Y \text{P-a.s} \iff E\{|X-Y|\}=0.$
So
$$E\{|X-Y|\}=E\{|2x-2x\cdot 1_{(0,1)}(x)|\}=\int 2x-2x\cdot 1_{(0,1)} f(x)dm(x)=$$ Since x is retricted to the interval $(0,1),$ then the indicator function in the integral equals 1 whichs gives: $$\int (2x-2x\cdot 1_{(0,1)}) 1_{(0,1)}(x)dm(x)=\int_{(0,1)} 2x-2x\cdot 1_{(0,1)}dm(x)=\int_{(0,1)} 2x-2xdm(x)=\int_{(0,1)} 0dm(x)=0$$
Is this correct?