Show $\mathcal{H}_\eta = L^2([a,b], \eta)$ is a Hilbert space when $\eta$ is positive, not necessarily continuous Exercise $8$ of Stein and Sharkarchi's Real Analysis asks first to show that the space of measurable $f$ on $[a,b]$ such that $$\int_a^b |f(t)|^2 \eta (t)dt < \infty $$ denotes $\mathcal{H}_\eta = L^2([a,b], \eta)$ is a Hilbert space with norm $$\langle f, g \rangle_{\mathcal{H}_\eta} = \int_a^b |f(t)\overline{g(t)}|^2\eta(t) dt.$$
A question about this first part of the exercise was asked here.
In the second part:

Generalize this to the case where $\eta$ is not necessarily continuous.

The key result of continuity that I used for the first part was that $\eta$ achieves both a minimum and maximum on $[a,b]$. However, this is obviously not the case if $\eta$ is only required to be strictly positive.
I know by a theorem of Lusin that since $$ \int_a^b |f(t)|^2\eta(t) < \infty $$ it is a.e. finite and we can find a closed subset of $[a,b]$ such that the integrand is continuous. However, this isn't proving to be very helpful because I can't think of how to split up $\eta$ and $f$ so I can consider them separately.
Any hints would be appreciated.
 A: Let $\| \cdot \|_\eta$ denote the norm on $\mathcal H_\eta$, then $\| \cdot \|_1$ is the usual $L^2$-norm ($\eta \equiv 1$). Because $\eta$ is positive $\sqrt \eta$ is real and thus $\| g \|_{\eta} = \| g \sqrt \eta \|_1$ for all $g \in \mathcal H_n$. This identity means that you can carry most results from $L^2$ to $\mathcal H_\eta$.
For example, to prove the completeness of $\mathcal H_\eta$: Let $(f_n)$ be a Cauchy sequence in $\mathcal H_\eta$, i.e. for $\epsilon > 0$ there is $N_\epsilon \in \mathbb N$ such that $\| f_n - f_m \|_\eta^2 \leq \epsilon$ for all $n, m \geq N$. This implies that $\| (f_n - f_m) \sqrt \eta \|_1^2 \leq \epsilon$ for all $n, m \geq N$. Hence $(f_n \sqrt \eta)$ is a Cauchy sequence in $L^2([a,b])$ and therefore, by completeness of $L^2$, there is $g \in L^2([a,b])$ s.t. $(f_n \sqrt \eta) \to g$ in $L^2([a,b])$. 
Because $\eta > 0$ we can set $f := \frac{g}{\sqrt \eta}$. Then $\| f \|_\eta = \| \frac{g}{\sqrt \eta} \|_\eta = \| g \|_1 < \infty$, i.e. $f \in \mathcal H_\eta$, and $\| f_n - f \|_\eta = \| (f_n - f) \sqrt \eta \|_1 = \| f_n \sqrt \eta - g \|_1 \to 0$ for $n \to \infty$, i.e. $f_n \to f$ in $\mathcal H_\eta$. 
