We know if $(x_{n})_{n=0}^{\infty}$ and $(y_{n})_{n=0}^{\infty}$ are real-valued sequences converging to $x$ and $y$ respectively and $x_{n}\leq y_{n}$, then $x<y$.
Now let $(\Omega, \Sigma,\mu)$ be a finite measure space. Suppose $x_{n}, y_{n}\in L^{2}(\Omega, \mu)$ with $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$ weakly. Suppose $x_{n}\leq y_{n}$ for $\mu$ almost everywhere. Do we have $x\leq y$ for $\mu$-a.e.?
Here is my attempt at a proof. Suppose by contradiction $x\leq y$ does not hold for $\mu$-a.e. By definition, we have, $\{\omega \in \Omega\vert x(\omega)>y(\omega)\}\in \Sigma$ and $\mu(\{\omega \in \Omega\vert x(\omega)>y(\omega)\})>0$.
Let $B \colon = \{\omega \in \Omega\vert x(\omega)>y(\omega)\}$. The indicator function $\mathbb{1}_{B}(\cdot)$ will be measurable; moreover, we have
\begin{equation} \int \mathbb{1}_{B}x \,d\mu >\int \mathbb{1}_{B}y\, d\mu \end{equation}
where strict inequalities are preserved according to an argument such as this.
However, since $x_{n} \leq y_{n}$ holds $\mu$-a.e., we must have (*)
\begin{equation} \int \mathbb{1}_{B}x_{n}\,d\mu \leq \int \mathbb{1}_{B}y_{n}\,d\mu \end{equation}
Since $\mathbb{1}_{B}\in L^{2}(\Omega,\mu)$, $\int\mathbb{1}_{B}x_{n}\,\mathrm{d}\mu \rightarrow \int\mathbb{1}_{B}x\,\mathrm{d}\mu$ and $\int\mathbb{1}_{B}y_{n}\,\mathrm{d}\mu \rightarrow \int\mathbb{1}_{B}y\,\mathrm{d}\mu$. And thus $\int\mathbb{1}_{B}x\,\mathrm{d}\mu \leq \int\mathbb{1}_{B}y\,\mathrm{d}\mu$, yielding a contradiction.
Is this reasoning correct? In particular, I am afraid of the step I take at *.
I would appreciate any suggestions/ corrections.
