Let $\omega(t)$ be a non-negative real Lebesgue-integrable function. $$H=\{f,\ \textrm{Lebesgue-measurable}: \int_{\mathbb{R}}|f(t)|^2\omega(t)dt<\infty\}$$ Prove that $H$ is a Hilbert space under the inner product $$\left<f,g\right>=\int_\mathbb{R}f(t)\overline{g(t)}\omega(t)dt, \forall f,g\in H$$
I can figure out that this inner product is well-defined and $H$ is a linear space. However, I don't know how to get the completeness of $H$. Any help would be appreciated! Thanks!
Updates: To check the completeness, I need to verify that every Cauchy sequence is convergent. First I get the induced norm of inner product: $$\|f\|=\sqrt{\int_\mathbb{R}|f(t)|^2\omega(t)dt}$$ Then I need to prove:
If $\{f_n\}\subset H$ satisfies $\forall \varepsilon>0$, $\exists N>0$, when $m>N$,
$$\|f_{m}-f_{m+p}\|<\varepsilon,(\forall p>0)$$
then there exists $f\in H$ such that $\|f_n-f\|\to 0(n\to \infty)$.
Here is where I am stuck.
Updates: I find this norm is similar to the norm on $L^2$, just multiplied a weight $\omega(t)$ at each $t$. so I guess $$\left<f,g\right>=\int_\mathbb{R}\left(f(t)\sqrt{\omega(t)}\right)\left(\overline{g(t)}\sqrt{\omega(t)}\right)dt=\left<f\sqrt{\omega},g\sqrt{\omega}\right>_{L^2}$$ would work. Then use the completeness of $L^2$.