change the order of $\limsup$ in sequence Let two positive function $u\in L^1$ and $v\in L^1$ so that 
$$
\int_\Omega uv=M<\infty
$$
where $\Omega\subset \mathbb R^N$ is open bounded. Define $v_k$ be the truncation of $v$ by positive constant $k>0$, i.e., $v_k(x)=v(x)$ if $v(x)<k$, o.w. $v_k(x)=k$. Suppose that I have a positive function sequence $u_n$ so that $u_n\to u$ in $L^1$ and
$$
\limsup_{n\to\infty} \int_\Omega u_nv_k\leq \int_\Omega uv_k\leq \int_\Omega uv<\infty
$$
for arbitrary $k>0$.
My question: can I conclude that 
$$
\limsup_{n\to\infty}\int_\Omega u_nv\leq \int_\Omega uv
$$

No, here's a counterexample: On $(0,1),$ let $u(x) = x^{-1/2}, v(x) = x^{-1/4}.$ Then $uv\in L^1(0,1).$ Let $u_n(x) = u(x) + x^{-3/4}/n.$ Then $u_n \to u$ in $L^1(0,1).$ However $\int_0^{1} u_nv = \infty$ for all $n.$
NOTE: I changed my previous answer; I think the above is easier.
