Suppose $X$ is a random variable with $EX=m$ and $V(X)=\sigma^2$. If $\sigma^2=0$, show that $P\{X=m\}=1$.
Following the Lebesgue integral definition of $V(X)$ and the property that if the integral of a non-negative integrand equals $0$ then the function must be $0$ a.e. wrt the measure. Then, $$\sigma^2=0\Rightarrow\int_\Omega(X(\omega)-m)^2P(d\omega)=0\Rightarrow P\{\omega:X(\omega)\ne m\}=0$$ Then, this implies that $P(X=m)=1$.
Is this solution correct? It seems too immediate a result to be asked a proof for. Are there some results or properties that I am missing in the proof I wrote or some extra steps I need in between?