Let $\Omega\subset \mathbb R^N$ open bounded. Given a sequence of Radon measure $(\mu_n)$ such that $\mu_n\to \mu$ in weak star sense in $\mathcal M_b(\Omega)$ and $\|\mu_n\|\nearrow \|\mu\|$. Also given a function $v\geq 1$ such that $v\in L^1_{\text{loc}}(\Omega)$ and $l.s.c$. Suppose $v$ has properties that there exists a Lipschitz continuous sequence $v_n$ such that $1\leq v_n\leq v$ and $v_n\nearrow v$ for all $x\in\Omega$. Note that $v$ may not bounded above.

Assume there exists a function $u\in C(\Omega)$ such that $u/v\in C_c(\Omega)$.

My question: do we have $$ \lim_{n\to\infty}\int_\Omega \frac{u}{v_n}\,d\mu_n = \int_\Omega \frac{u}{v}\,d\mu $$

My try:

Writing $$ \int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu = \int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu_n+\int_\Omega \frac{u}{v}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu $$ The last two goes to 0 by the definition of weak star convergence. But I don't know how to deal with first two. I was trying to use dominated convergence but it is not obvious...

I realize that since $u/v\in C_c(\Omega)$ and $v$ is finite a.e., it makes $u$ has compact support and hence $u$, $u/v_n\in C_c(\Omega)$ since $u\in C(\Omega)$. Now, I could write few more steps... \begin{align*} \left|\int_\Omega \frac{u}{v_n}\,d\mu_n - \int_\Omega \frac{u}{v}\,d\mu_n\right| &= \left|\int_\Omega u(1/v_n-1/v)\,d\mu_n\right|\\ &=\left|\int_\Omega u(1/v_n-1/v)\,d(\mu_n-\mu+\mu)\right|\\ &\leq \left|\int_\Omega u(1/v_n-1/v)\,d(\mu_n-\mu)\right|+\left|\int_\Omega u(1/v_n-1/v)\,d\mu\right| \end{align*} The last one can be done by dominated convergence. But the first one...Maybe I should use the fact that $\|\mu_n\|\to\|\mu\|$?

I know generally I should not hope $$ \lim_{n\to\infty}\int_\Omega \frac{u}{v_n}\,d\mu_n = \int_\Omega \frac{u}{v}\,d\mu $$ since it would require that $u/v_n\to u/v$ uniformly which I don't have. But since I in additional have $0\leq 1/v\leq 1/v_n\leq 1$ and $\|\mu_n\|\to \|\mu\|$, I may expect my result is true.

  • $\begingroup$ You state that $u/v_n$ may not be in $C_c$, but since $u/v$ has compact support, so has $u=v \cdot u/v$, so that we get $u/v_n$ is also $C_c$, right? $\endgroup$ – PhoemueX Aug 18 '15 at 9:20
  • $\begingroup$ @PhoemueX you are right sir. I updated my post. Please have a look. $\endgroup$ – spatially Aug 18 '15 at 13:06

We will use the version of Fatou's Lemma given here (https://en.wikipedia.org/wiki/Fatou%27s_lemma#Fatou.27s_Lemma_with_Varying_Measures). This states that if we have $\mu_n (A) \to \mu(A)$ for all measurable $A$ and if $(f_n)_n$ is a sequence of nonnegative measurable functions, then $$ \int \liminf_n f_n \, d\mu \leq \liminf_n \int f_n \, d\mu_n . $$ In your case, this is satisfied, since you assume $\mu_n \to \mu$ weak star with respect to $M_b$ (and every indicator function is in $M_b$).

But this statement implies a nice version of the dominated convergence theorem: If we have $|f_n| \leq g$ with $\int g \, d\mu_n \to \int g \, d\mu< \infty$ (for example for every bounded measurable function $g$) and $f_n \to f$ pointwise, then $\int f_n \, d\mu_n \to \int f \, d\mu$. Indeed, let $h_n := 2g + |f_n -f|$. Then $h_n \geq 0$ and $h_n \to 2g$ pointwise, so that Fatou's lemma from above yields $$ \int 2g \, d\mu \leq \liminf_n \int h_n \, d\mu_n = \int 2g \, d\mu - \limsup_n \int |f_n - f| \, d\mu_n $$ and thus $$ \lim_n \int |f_n - f|\, d\mu_n = 0. $$ Furthermore, if we apply Fatou's lemma with $h_n = g \pm f$ (note $h_n \geq 0$), we get $$ \int g \pm f \, d\mu \leq \liminf_n \int g \pm f \, d\mu_n = \int g \, d\mu \pm \lim^\ast \int f_n \, d\mu_n, $$ with $\lim^\ast = \liminf$ for $\pm = +$ and $\lim^\ast = \limsup$ for $\pm = -$. Hence, $$ \int f \, d\mu \leq \liminf \int f_n \, d\mu_n \leq \limsup \int f_n \, d\mu_n \leq \int f \, d\mu, $$ i.e. $\inf f_n \, d\mu_n \to \int \int f \, d\mu$. Together with the above, this implies $$ \int f_n \, d\mu_n \to \int f\, d\mu. $$ This is the version of the dominated convergence theorem that we will use.

Now, it remains to note (since $v_n \geq 1$) that $$ \bigg| \frac{u}{v_n}\bigg| \leq \Vert u \Vert_\sup \cdot \chi_K =: g, $$ where $K$ is the support of $u$. Since all $\mu_n$ are Radon, we have $\int g \, d\mu_n \to \int g \, d\mu < \infty$ ($g$ is bounded measurable). Furthermore, we have pointwise convergence. Now apply the dominated convergence theorem from above.

  • $\begingroup$ Nice proof! Indeed, this version of LOCT is extremely good. Also, from your argument, I would say this result still hold for $\mu_n$ and $\mu$ are signed Radon measure right? Since by $\|\mu_n\|(\Omega)\to \|\mu\|(\Omega)$ we have the total variation measure $|\mu_n|\to |\mu|$ in weak star sense as well, and we can use Jorden decomposition. $\endgroup$ – spatially Aug 18 '15 at 19:24

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