I have to solve one problem about uniformly integrability and tightness. It is
"Let $f$ be a continuous integrable function on $\mathbb{R}$, and let $g$ be a bounded measurable function on $\mathbb{R}$. For each $n \in \mathbb{N}$, let $f_n(x) = f(x+1/n)$.
Prove $\{g \cdot f_n\}_{n \in \mathbb{N}}$ is uniformly integrable and tight."
I use Real Analysis of Royden as the textbook. The definitions of uniformly integrability and tightness as follows.
Uniformly integrable:
A family $\mathcal{F}$ of measurable functions on $E$ is said to be uniformly integrable over $E$ if $\forall \varepsilon > 0, \exists \delta > 0$ such that $\forall f \in \mathcal{F}$, if $A \subset E$ is measurable and $m(A) < \delta$, then $\int_A |f| < \varepsilon$.
Tightness:
A family $\mathcal{F}$ is said to be tight on $E$ if $\forall \varepsilon > 0$, there exists a set $E_0 \subset E$ such that $m(E_0) < \infty$ and $\forall f \in \mathcal{F}$,
$$\int_{E \setminus E_0} |f| < \varepsilon.$$
Here is my attempt:
- Uniformly integrability
Because $g$ is bounded on $\mathbb{R}$, there exists $M>0$ such that $|g| \le M$ on $\mathbb{R}$. Therefore, for any set $A \subset \mathbb{R}$, we have
$$ \int_{A} |gf_n| \le M \int_A |f_n| = M \int_A |f(x+1/n)|$$
I think I will evaluate the integral $f(x+1/n)$ on $A$ when $m(A) < \delta$ by some way, I don't know how to use the hypothesis "f is continous on $\mathbb{R}$". I think it will help me solve it.
- Tightness
Sadly, I totally got stuck in this.
Please give me some hints. I think they are about the continuity of $f$ and the boundedness of $g$. Any help is highly appreciated.
Best Answer
Since $\int_A |f(x+1/n)|=\int_{A+1/n} |f(x)|$, where $A+1/n=\{a+1/n,a\in A\}$, uniform integrability follows from the fact that $\lim_{\delta\to 0}\sup_{A:\mu(A)\leq\delta}\int_A\lvert f(x)\rvert dx$.
For the tightness part, write for a set $E_0$ of finite measure that will be specified later, $$ \int_{E \setminus E_0} |(gf_n)(x)| \leqslant \int_{E \setminus E_0} |f_n(x)| =\int_{(E\setminus E_0)+1/n}\lvert f(x)\rvert $$ so concretely, it suffices to take $E_0$ of the form $[-R,R]$ where $R$ is such that $\int_{\mathbb R\setminus [-R+1,R-1]}\lvert f(x)\rvert\lt\varepsilon$.