Is it true that
$$\int_{[0,1]}\int_{[0,1]} f(t)g(s)2^m\sum_{\substack{i=0}}^{2^{m}-1}\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(t)\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(s)\mathrm{d}t\mathrm{d}s\xrightarrow{m\to\infty}\int_{[0,1]}f(t)g(t)\mathrm{d}t$$ holds, where $f,g$ are square-integrable?
This comes from trying to find a representation for time white noise.
I feel like it should because as $m$ goes to infinity, the indicator functions should `force' $s$ and $t$ to become equal and the length of that interval would be $\frac{1}{2^m}$.
This resembles the definition of the delta distribution: if we were to substitute the part of the integrand which isn't $f(t)g(s)$ by $\delta(t-s)$, we would get $\int_{[0,1]}f(t)g(t)\mathrm{d}t$.
Any hints/suggestions on how can I prove this rigorously (if it even holds)?
EDIT: some details for where this comes from:
Let's start with this representation for time white noise: $$\xi(t)=\lim_{m\to\infty}2^{\frac{m}{2}}\sum_{i=0}^{2^m-1}X_{i,m}\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(t),$$ where the $X_{i,m}$ are i.i.d. standard normal random variables. We need to show that it satisfies the properties to be time white noise, that is to say: $$\mathbb{E}[\xi(f)\xi(g)]=\langle f,g\rangle.$$ The left hand side equals (using the independence of the random variables considered) \begin{align*}&\mathbb{E}\left[\int_0^1\xi(t)f(t)\mathrm{d}t\int_0^1\xi(t)g(t)\mathrm{d}t\right]\\ &=\mathbb{E}\left[\int_0^1\xi(t)f(t)\mathrm{d}t\right]\mathbb{E}\left[\int_0^1\xi(t)g(t)\mathrm{d}t\right]\\ &=\int_0^1\mathbb{E}\left[\xi(t)\right]f(t)\mathrm{d}t\int_0^1\mathbb{E}\left[\xi(t)\right]g(t)\mathrm{d}t\mbox{ (still not sure how I can justify this)}\\ &=\int_0^1\int_0^1 f(t)g(s)\mathbb{E}\left[\xi(s)\xi(t)\right]\mathrm{d}t\mathrm{d}s\\ &=\int_0^1\int_0^1 f(t)g(s)\mathbb{E}\left[\lim_{\substack{m\to\infty\\n\to\infty}}2^{\frac{m}{2}+\frac{n}{2}}\sum_{\substack{i=0\\j=0}}^{\substack{2^{m}-1\\2^n-1}}X_{i,m}X_{j,n}\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(t)\mathbf{1}_{\left[\frac{j}{2^n},\frac{j+1}{2^n} \right]}(t)\right]\mathrm{d}t\mathrm{d}s\\ &=\int_0^1\int_0^1 f(t)g(s)\lim_{\substack{m\to\infty\\n\to\infty}}2^{\frac{m}{2}+\frac{n}{2}}\sum_{\substack{i=0\\j=0}}^{\substack{2^{m}-1\\2^n-1}}\mathbb{E}\left[X_{i,m}X_{j,n}\right]\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(t)\mathbf{1}_{\left[\frac{j}{2^n},\frac{j+1}{2^n} \right]}(t)\mathrm{d}t\mathrm{d}s\\ &=\int_0^1\int_0^1 f(t)g(s)\lim_{\substack{m\to\infty}}2^{m}\sum_{\substack{i=0}}^{\substack{2^{m}-1}}\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(t)\mathbf{1}_{\left[\frac{i}{2^m},\frac{i+1}{2^m} \right]}(s)\mathrm{d}t\mathrm{d}s. \end{align*}
Write $I_{m,i} = [\frac{i}{2^m},\frac{i+1}{2^m}]$ and introduce the averaging operator $A_m : L^2([0,1]) \to L^2([0,1])$ given by
$$ (A_m f)(t) = \sum_{i=0}^{2^m-1} \left( 2^m \int_{I_{m,i}} f(s) \, \mathrm{d}s \right) \mathbf{1}_{I_{m,i}}(t). $$
We claim that $A_m$ is a bounded operator on $L^2([0,1])$. Indeed, this follows from the following computation
\begin{align*} \|A_m f(t)\|_{L^2}^2 &= \int_{0}^{1} |A_m f(t)|^2 \, \mathrm{d}t \\ &\leq \int_{0}^{1} \sum_{i=0}^{2^m-1} \left( 2^m \int_{I_{m,i}} |f(s)| \, \mathrm{d}s \right)^2 \mathbf{1}_{I_{m,i}}(t) \, \mathrm{d}t \\ &= \sum_{i=0}^{2^m-1} 2^m \left( \int_{I_{m,i}} |f(s)| \, \mathrm{d}s \right)^2 \\ &\stackrel{\text{C-S}}{\leq} \sum_{i=0}^{2^m-1} \int_{I_{m,i}} |f(s)|^2 \, \mathrm{d}s = \|f\|_{L^2}^2. \end{align*}
Notice that this operator is related to the quantity of interest by the following identity:
$$ \int_{[0,1]}\int_{[0,1]} f(t)g(s) 2^m \sum_{i=1}^{2^m-1}\mathbf{1}_{I_{m,i}}(t)\mathbf{1}_{I_{m,i}}(s) \, \mathrm{d}t\mathrm{d}s = ( A_m f, g )_{L^2} = ( f, A_m g )_{L^2} $$
Next, if $f$ is continuous then it is easy to prove that $A_m f \to f$ in $L^{\infty}$ and hence in $L^2$. Thus for any $f, g \in L^2([0,1])$ and for any $\tilde{g} \in C([0,1])$, we have
\begin{align*} |(f, A_m g)_{L^2} - (f, g)_{L^2}| &\leq \|f\|_{L^2} \| A_m g - g\|_{L^2} \\ &\leq \|f\|_{L^2} \left( \| A_m(g - \tilde{g}) \|_{L^2} + \| A_m \tilde{g} - \tilde{g} \|_{L^2} + \| \tilde{g} - g \|_{L^2} \right) \\ &\leq \|f\|_{L^2} \left( 2 \| g - \tilde{g} \|_{L^2} + \| A_m \tilde{g} - \tilde{g} \|_{L^2} \right). \end{align*}
Taking limsup as $m\to\infty$, since $A_m \tilde{g} \to \tilde{g}$ in $L^2$, we have
$$ \limsup_{m\to\infty} |(f, A_m g)_{L^2} - (f, g)_{L^2}| \leq 2 \|f\|_{L^2} \| g - \tilde{g} \|_{L^2}. $$
Then taking $\tilde{g} \to g$ in $L^2$ proves the desired conclusion:
$$ \lim_{m\to\infty} (f, A_m g)_{L^2} = (f, g)_{L^2}. $$
Remark. This result itself tells that $A_m \to I$ in the weak operator topology. On the other hand, a slight modification of the proof actually tells that $A_m \to I$ in the strong operator topology:
$$ \forall g \in L^2([0,1]) \ : \qquad A_m g \to g \quad \text{in } L^2. $$
I am not sure if $A_m \to I$ in the operator-norm topology.