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.


Your proof is correct, including (*). Since $x_n(x)\le y_n(x)$ for all $x\in \Omega\setminus E$ where $\mu(E)=0$, you have that $0\le y_n(x)-x_n(x)$ for all $x\in B \setminus E$. If you integrate over $B\setminus E$ you get $0\le\int_{B\setminus E}(y_n-x_n)\,d\mu$ since the integral of a nonnegative function is a nonnegative number. Since $E$ has measure zero (so the integral does not see it) you get $\int_Bx_n\,d\mu\le\int_By_n\,d\mu$.


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