Convergence of sequence of integrals.

Question

Let $(\mathcal{X}, \mathcal{A}, \mu)$ be a measure space, $f_n: \mathcal{X} \to \Bbb R$ a sequence of measurable functions, and $g_n:\mathcal{X} \to \Bbb R$ integrable functions such that $|f_n| \leq g_n$. Let $f$ and $g$ measurable functions, $g$ integrable, with $f_n(x) \to f(x)$ and $g_n(x) \to g(x)$. If $\int_{\mathcal{X}} g_n \ \mathrm{d}\mu \to \int_{\mathcal{X}} g \ \mathrm{d}\mu $, then $\int_{\mathcal{X}} f_n \ \mathrm{d}\mu \to \int_{\mathcal{X}} f \ \mathrm{d}\mu $.

I think that the idea is to dominate all those $f_n$ with some integrable function, and then by the Dominated Convergence Theorem, we're done. The problem is that each $f_n$ is dominated by $g_n$, it's not the same function dominating all of them. Ok, I know that: $$f_n \leq |f_n(x)| \leq g_n(x) \implies |f(x)| \leq g(x) \quad \mbox{and}\quad \int_{\mathcal{X}} f_n \ \mathrm{d}\mu \leq \int_{\mathcal{X}} g_n \ \mathrm{d}\mu.$$

Taking limits, we get: $$\limsup \int_{\mathcal{X}} f_n \ \mathrm{d}\mu \leq \int_{\mathcal{X}} g \ \mathrm{d}\mu $$

But other than this, I don't know how to do it. Can someone give me some help (or hints of how to use this info)? Thanks.

Answer 1

HINT: $h_n = f_n + g_n$ and $\widehat{h_n} = g_n - f_n$ and apply Fatou's Lemma to $\int \liminf h_n$ and $\int \liminf \widehat{h_n}$

Solution: Let $h = f+g$, $h_n$ and $\widehat{h_n}$ as before, because $- g_n \leq f_n \leq g_n$ we have that both $h_n$ and $\widehat{h_n}$ are greater or equal than $0$. Now:

$$\int f + g = \int h =\int \liminf h_n \leq \liminf \int h_n = \liminf \int f_n + g_n$$ $$\leq \liminf \int f_n + \limsup \int g_n = \liminf \int f_n + \int g$$

So $\int f \leq \liminf \int f_n$

$$\int \liminf \widehat{h_n} \leq \liminf \int h_n \leq \limsup \int g_n + \liminf (- \int f_n) $$

Using properties of limsup ($\liminf - \int f_n = - \limsup \int f_n$)

$$= \limsup \int h_n - \limsup \int f_n = \int g - \limsup \int f_n$$

So $\int f \geq \limsup \int f_n$

Answer 2

Let $(\mathcal{X}, \mathcal{A}, \mu)$ be a measure space, $f_n: \mathcal{X} \to \Bbb R$ a sequence of measurable functions, and $g_n:\mathcal{X} \to \Bbb R$ integrable functions such that $|f_n| \leq g_n$. Let $f$ and $g$ measurable functions, $g$ integrable, with $f_n(x) \to f(x)$ and $g_n(x) \to g(x)$. If $\int_{\mathcal{X}} g_n \ \mathrm{d}\mu \to \int_{\mathcal{X}} g \ \mathrm{d}\mu $, then $\int_{\mathcal{X}} f_n \ \mathrm{d}\mu \to \int_{\mathcal{X}} f \ \mathrm{d}\mu $.

First, we show that $f$ is integrable. Since $f_n \rightarrow f$, it follows that $|f_n|\rightarrow |f|$. Since for each $n$, $|f_n|\leq g_n$, it follows that $|f|\leq |g|$. |g| is integrable, so is |f|. But then $f$ is integrable, as desired.

Now notice that for each $n$, $-g_n \leq f_n \leq g_n$. So for each $n$, $0 \leq f_n+g_n$. By the Fatou's Lemma, it follows that $$\int_Xfd \mu+\int_Xgd \mu=\int_X(f+g)d \mu\leq \lim \inf \int_X(f_n+g_n)d \mu=\lim \inf( \int_Xf_n d\mu+\int_X g_n d\mu)=\lim \inf(\int_X f_n d \mu)+\int_Xg d\mu$$ Therefore, $\int_Xfd \mu\leq\lim \inf(\int_X f_n d \mu)$

Notice that $f+g$ may not be defined in all $X$ since we may have $f(x)=\infty$ and $g(x)=-\infty$ or $f(x)=-\infty$ and $g(x)=\infty$. But since $f$ is integrable, this is false almost everywhere, so you may just redefine $f$ so that $f=0$ where it was $+\infty$.

Now notie that for each $n$, $0 \leq g_n-f_n$. Again, as above:

$$\int_Xgd\mu-\int_Xfd\mu=\int_Xg-fd\mu\leq\lim\inf\int_X g_n-f_n d \mu = \lim \inf(\int_X g_n d \mu - \int_X f_n d \mu)=\int_X g d\mu+\lim \inf(-\int_X f_n d \mu)=\int_X g d\mu-\lim \sup(\int_X f_n d \mu)$$

Therefore: $\lim \sup \int_X f_n d \mu \leq \int f d \mu$

We have obtained: $$\int_X f d \mu \leq \lim \inf\int f_n d \mu\leq\lim \sup \int_X f_n d \mu \leq \int f d \mu$$ So the equality holds and $\lim_{n \rightarrow \infty}\int_X f_n d \mu = \int_X f d \mu$.