Calderon-Zygmund Operator Associated to Zero Kernel We say that a Calderon-Zygmund operator (CZO) $T:L^{2}(\mathbb{R}^{n})\rightarrow L^{2}(\mathbb{R}^{n})$ (i.e. a bounded linear operator) is associated to a CZ kernel $K:\mathbb{R}^{n}\times\mathbb{R}^{n}\setminus\Delta$ ($\Delta$ denotes the diagonal subset) if the following holds: for $f\in L_{c}^{2}(\mathbb{R}^{n})$ fixed, for almost everywhere (a.e.) $x\notin\text{supp}(f)$,
    $$Tf(x)=\int_{\mathbb{R}^{n}}K(x,y)f(y)dy$$
Suppose that a CZO $T:L^{2}(\mathbb{R}^{n})\rightarrow L^{2}(\mathbb{R}^{n})$ is associated the kernel $K(x,y)=0$. In the hint for Question 2 in the Exercises section of Tao's Lecture Notes 4, he suggests that we can show $T$ is given by pointwise multiplication by a bounded function (i.e. $Tf=bf$, where $b\in L^{\infty}$) by observing that the set function
    $$E\mapsto \langle{T(\chi_{E}),\chi_{E}}\rangle$$
is an absolutely continuous measure and then applying the Radon-Nikodym theorem. I see where the Radon-Nikodym theorem gives us a measurable function $b$ such that
    $$\langle{Tf,f}\rangle=\int_{\mathbb{R}^{n}}b\left|f\right|^{2},\quad\forall f\in L^{2}(\mathbb{R}^{n})$$
by the boundedness of $T$, the density of simple functions in $L^{2}$, and the linearity of the integral. But I am failing to see (perhaps very foolishly) why we can readily deduce that $Tf=bf$ for all $f\in L^{2}$. I believe I can obtain the desired result using a different argument, as shown below; but I would still like to see how to make Tao's suggestion work. Does anyone have a suggestion?

Let $g\in L^{2}(\mathbb{R}^{n})$ have compact support. For any cube $Q\subset\mathbb{R}^{n}$,
    $$T(\chi_{Q}g)(x)=\int_{\mathbb{R}^{n}}K(x,y)\chi_{Q}(y)g(y)dy=0 \quad \text{ a.e. } x\notin\overline{Q}$$
and
    $$T(\chi_{Q^{c}}g)(x)=\int_{\mathbb{R}^{n}}K(x,y)\chi_{Q^{c}}(y)g(y)dy=0 \quad \text{ a.e. } x\in Q^{o}$$
Since $T(\chi_{Q}g)=T(g)-T(\chi_{Q^{c}}g) \text{ a.e. }$, we conclude that
    $$T(\chi_{Q}g)(x)=\chi_{Q}(x)(Tg)(x) \quad \text{ a.e. }$$
Since the collection of all cubes with rational vertices is countable and finite linear combinations of the characteristic functions of such cubes are dense in $L^{2}$, the boundedness of $T$ yields
    $$T(fg)=f(Tg) \quad\text{ a.e. },\qquad\forall f\in L^{2}(\mathbb{R}^{n})$$
To see that $T$ is given by pointwise multiplication by a bounded function $b$, let $B_{j}$ be the ball centered at the origin of radius $j$. Observe that if $j'\geq j$, then $\chi_{B_{j}}\chi_{B_{j'}}=\chi_{B_{j}}$ and therefore
    $$T(\chi_{B_{j}})=T(\chi_{B_{j}}\chi_{B_{j'}})=\chi_{B_{j}}T(B_{j'}) \quad \text{ a.e. }$$
by our previous result. Thus, we can define almost everywhere a measurable function $b$ by
    $$b(x)=T(\chi_{B_{j}})(x), \quad x\in B_{j}$$
For an $f\in L^{2}(\mathbb{R}^{n})$, let $f_{j}=f\chi_{B_{j}}$.Then by the continuity of $T$ and a.e. convergence $f_{j}\rightarrow f$
    $$T(f)=\lim_{j\rightarrow\infty}T(f_{j})=\lim_{j\rightarrow\infty}bf_{j}=bf \quad \text{ a.e. }$$
To see that $b$ is bounded, observe that
    $$\left\|T\right\|_{L^{2}\rightarrow L^{2}}=\sup_{f\in L^{2}}\left\|Tf\right\|_{L^{2}}=\sup_{f\in L_{c}^{2}}\left\|bf\right\|_{L^{2}}=\left\|b\right\|_{L^{\infty}}$$
 A: By using the first part of the alternative solution I gave in my question post, we have that
$$T(\chi_{E\cap F})=T(\chi_{E}\chi_{F})=\chi_{F}T(\chi_{E}) \quad \text{a.e.}$$
for all measurable sets $E$ and $F$ with $\left|E\right|,\left|F\right|<\infty$. Following the hint, we can show that
$$\langle{T(\chi_{A}),\chi_{A}}\rangle=\int b\chi_{A},\quad\forall A\subset\mathbb{R}^{n},\left|A\right|<\infty$$
By Lebesgue differentiation theorem,
\begin{align*}
\left|b(x)\right|\leq\limsup_{{Q\ni x}\atop{\left|Q\right|\rightarrow 0}}\left|Q\right|^{-1}\left|\int_{Q}b\right|&\leq\limsup_{{Q\ni x}\atop{\left|Q\right|\rightarrow 0}}\left|Q\right|^{-1}\left\|T\chi_{Q}\right\|_{L^{2}}\left|Q\right|^{1/2}\\
&\leq\left\|T\right\|_{L^{2}\rightarrow L^{2}}\quad \text{a.e.}
\end{align*}
by Cauchy-Schwarz. Whence, $b\in L^{\infty}(\mathbb{R}^{n})$.
Whence,
$$\langle{T\chi_{E},\chi_{F}}\rangle=\int_{E\cap F}b=\int b\chi_{E}\chi_{F}$$
By density of simple functions, linearity, and $L^{2}$ boundedness of $T$, we conclude that
$$\langle{Tf,g}\rangle=\langle{bf,g}\rangle,\quad\forall f,g\in L^{2}(\mathbb{R}),$$
which implies that $Tf=bf$ a.e. 
