2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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, $\limsup_n\int_Xf_nd\mu\leq\int_X\limsup_nf_nd\mu$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.