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First, if $uv=0$ then $uTv=0$ a.e.: indeed, applying the hypothesis with the function $\epsilon u+\epsilon^{-1}v$ and evaluating on $\{u\neq 0\}$, we get $$0\le \epsilon^2uTu+\epsilon^{-2}vTv+uTv+vTu=\epsilon^2 uTu+uTv$$ a.e. and deduce the claim (sending $\epsilon\to 0$). This gives the property $(*)$ that you showed.

If the space is $\sigma$-finite, using $(*)$ we can easily reduce to the case that the measure is finite. Now take $v:=T1$. Since $T$ is linear and simple functions are dense, it suffices to show that $Tu=uv$ (a.e.) when $u=1_A$ is a characteristic function. But $$v=T1=T1_A+T1_{A^c}$$ and $T1_{A^c}$ vanishes (a.e.) on $A$, hence $T1_A=v$ (a.e.) on $A$. Since $T1_A$ vanishes (a.e.) on $A^c$, the claim follows.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.

First, if $uv=0$ then $uTv=0$ a.e.: indeed, applying the hypothesis with the function $\epsilon u+\epsilon^{-1}v$ and evaluating on $\{u\neq 0\}$, we get $$0\le \epsilon^2uTu+\epsilon^{-2}vTv+uTv+vTu=\epsilon^2 uTu+uTv$$ a.e. and deduce the claim (sending $\epsilon\to 0$). This gives the property $(*)$ that you showed.

If the space is $\sigma$-finite, using $(*)$ we can easily reduce to the case that the measure is finite. Now take $v:=T1$. We claim that $Tu=uv$ (a.e.) when $u$ is simple: since $T$ is linear, it suffices to show this when $u=1_A$ is a characteristic function. But $$v=T1=T1_A+T1_{A^c}$$ and $T1_{A^c}$ vanishes (a.e.) on $A$, hence $T1_A=v$ (a.e.) on $A$. Since $T1_A$ vanishes (a.e.) on $A^c$, the claim follows. Taking $A:=\{|v|>\|T\|\}$, if $\mu(A)>0$ we see that $\|T1_A\|_{L^2}>\|T\|\|1_A\|_{L^2}$, contradiction. So $v\in L^\infty$. The statement now follows since simple functions are dense.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.

First, if $uv=0$ then $uTv=0$ a.e.: indeed, applying the hypothesis with the function $\epsilon u+\epsilon^{-1}v$ and evaluating on $\{u\neq 0\}$, we get $$0\le \epsilon^2uTu+\epsilon^{-2}vTv+uTv+vTu=\epsilon^2 uTu+uTv$$ a.e. and deduce the claim (sending $\epsilon\to 0$). This gives the property $(*)$ that you showed.

If the space is $\sigma$-finite, using $(*)$ we can easily reduce to the case that the measure is finite. Now take $v:=T1$. Since $T$ is linear and simple functions are dense, it suffices to show that $Tu=uv$ (a.e.) when $u=1_A$ is a characteristic function. But $$v=T1=T1_A+T1_{A^c}$$ and $T1_{A^c}$ vanishes (a.e.) on $A$, hence $T1_A=v$ (a.e.) on $A$. Since $T1_A$ vanishes (a.e.) on $A^c$, the claim follows.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E$ and $E'\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$.

It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form $$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$ where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form $$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$ where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$. Once you have this, you can argue as before.