# A version of reverse Fatou's Lemma

Let $(X,M,\mu)$ be measure space and let $g_n:X\to[0,\infty]$ be a sequence of measurable functions such that $\int_Xg_nd\mu<\infty$. Suppose $\exists g:X\to[0,\infty]$ such that $\lim_n\int_X|g-g_n|d\mu=0$. Let $f_n:X\to[0,\infty]$ be another sequence of measurable functions such that $f_n(x)\leq g_n(x)$ almost everywhere. Prove $\limsup_n\int_Xf_nd\mu\leq\int_X\limsup_nf_nd\mu$.

The proof will be immediate using dominated convergence theorem if there exists an upper bound for $f_n$ (or $h_k=\sup_{n\geq k}f_n$), but the information about $g_n$ and $g$ given in the assumption is a little complicated, can we found such upper bound?

Since $0\leq g=g-g_n+g_n\leq|g-g_n|+g_n$, we have $0\leq\int_Xgd\mu\leq\int_X|g-g_n|d\mu+\int_Xg_nd\mu$. Since $\lim_n\int_X|g-g_n|d\mu=0$ and $\int_Xg_nd\mu<\infty$ for each $n$, so for $n$ large, $\int_Xgd\mu<\infty$. Next, since $|g_n-g|+g-f_n\geq g_n-g+g-f_n=g_n-f_n\geq0$ almost everywhere, by ignoring these points, we may apply [Rudin 2] 1.28 (Fatou's Lemma) on $|g_n-g|+g-f_n$, thus $\int_X\liminf_n(|g_n-g|+g-f_n)d\mu\leq\liminf_n\int_X(|g_n-g|+g-f_n)d\mu$. Then $\int_X\liminf_n|g_n-g|d\mu+\int_Xgd\mu-\int_X\limsup_nf_nd\mu=\int_X(\liminf_n|g_n-g|+\liminf_n\int_Xg+\liminf_n(-f_n))d\mu\leq\int_X\liminf_n(|g_n-g|+g-f_n)d\mu\leq\liminf_n\int_X(|g_n-g|+g-f_n)d\mu\leq\limsup_n\int_X|g_n-g|d\mu+\limsup_n\int_Xgd\mu+\liminf_n\int_X(-f_n)d\mu=\limsup_n\int_X|g_n-g|d\mu+\int_Xgd\mu-\limsup_n\int_Xf_nd\mu$, which implies $\int_X\liminf_n|g_n-g|d\mu-\int_X\limsup_nf_nd\mu\leq\limsup_n\int_X|g_n-g|d\mu-\limsup_n\int_Xf_nd\mu$. On the other hand, applying [Rudin 2] 1.28 (Fatou's Lemma) on $|g_n-g|$, we have $0\leq\int_X\liminf_n|g_n-g|d\mu\leq\liminf_n\int_X|g_n-g|d\mu=\limsup_n\int_X|g_n-g|d\mu=\lim_n\int_X|g_n-g|d\mu=0$. Therefore, $\limsup_n\int_Xf_nd\mu\leq\int_X\limsup_nf_nd\mu$.