interchanging sup and integral

Question

I have the following question:

Let $f_n:[0,1]\to [0,\infty)$ be Lebesgue measurable and such that $f_n(x)\to 0$ for almost every $x$. Assume further that \begin{equation} \sup_n\int_0^1 g(f_n(x))dx\leq 1 \end{equation} for some continuous $g:[0,\infty)\to[0,\infty)$ that satisfies $g(x)/x\to \infty$ as $x\to\infty$. Show that $\lim_{n}\int_0^1f_n(x)dx=0$.

My attempt: Since $f_n\to 0$ a.e., it suffices to show that we can interchange the limit and the integral in $\lim\int f_n(x)dx$. To that end, we seek $h(x)$ that is integrable and $|f(x)|\leq |h(x)|$ for a.e. $x$.

Since $g(x)/x\to\infty$ as $x\to\infty$, there exists some $M>0$ such that for all $x\geq M$, $g(x)>x$. Define $h_n(x):[0,1]\to[0,\infty)$ to be a piecewise function such that: \begin{equation} h_n(x) = \left\{\begin{array}{ll} M & \text{if $f_n(x)<M$,}\\ g(f_n(x)) & \text{if $f_n(x)\geq M$}. \end{array}\right. \end{equation} Clearly, $|f_n(x)|\leq |h_n(x)|$ for every $n$. Our objective is now to find one $h(x)$ that works for all $n$, and this is where I am stuck.

To show integrability of each $h_n(x)$, we observe that \begin{equation}\begin{split} \int_0^1 h_n(x)dx & = \int_{\{f_n(x)<M\}}Mdx+\int_{\{f_n(x)\geq M\}}g(f_n(x))dx\\ & \leq M + \int_0^1 g(f_n(x))dx \leq M+1 \end{split}\end{equation}

My guess is that to find one $h(x)$ that works for all $n$, we should take $\sup_n h_n(x)$, but then I don't know if \begin{equation} \int_0^1 \sup_n h_n(x)dx =(?) \sup_n \int_0^1 h_n(x)dx. \end{equation}

Any hint will be greatly appreciated!

Answer 1

Given $\epsilon>0$, find some $N>0$ such that \begin{align*} \dfrac{g(x)}{x}\geq\dfrac{1}{\epsilon},~~~~x\geq N, \end{align*} then \begin{align*} \int_{0}^{1}f_{n}(x)dx&=\int_{(f_{n}<N)}f_{n}(x)dx+\int_{(f_{n}\geq N)}f_{n}(x)dx. \end{align*} While, \begin{align*} \int_{(f_{n}\geq N)}f_{n}(x)dx&=\int_{(f_{n}\geq N)}\dfrac{f_{n}(x)}{g(f_{n}(x))}\cdot g(f_{n}(x))dx\\ &\leq\int_{(f_{n}\geq N)}\epsilon\cdot g(f_{n}(x))dx\\ &\leq\epsilon\cdot\sup_{n}\int_{0}^{1}g(f_{n}(x))dx\\ &\leq\epsilon. \end{align*} On the other hand, since $f_{n}(x)\rightarrow 0$ a.e., by Egorov Theorem, we can find a measurable set $S$ such that $|S|<\epsilon/N$ such that $f_{n}\rightarrow 0$ uniformly on $[0,1]-S$. As a result, \begin{align*} \int_{(f_{n}<N)}f_{n}(x)dx&=\int_{(f_{n}<N)\cap S}f_{n}(x)dx+\int_{(f_{n}<N)\cap([0,1]-S)}f_{n}(x)dx\\ &\leq N\cdot|S|+\int_{[0,1]-S}f_{n}(x)dx\\ &\leq\epsilon+\int_{[0,1]-S}f_{n}(x)dx. \end{align*} The uniform convergence of $f_{n}$ on $[0,1]-S$ entails that \begin{align*} \int_{[0,1]-S}f_{n}(x)dx\rightarrow 0, \end{align*} so \begin{align*} \limsup_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x)dx\leq 2\epsilon. \end{align*} The arbitrariness of $\epsilon>0$ gives that \begin{align*} \lim_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x)dx=0. \end{align*}

Answer 2

Let $\epsilon >0$. Choose $M$ such that $\frac t {g(t)} <\epsilon $ for $t >M$. Then $\int_{f_n >M} f_n(t)dt =\int_{f_n >M} g(f_n(t)) \frac {f_n(t)} {g(f_n(t))} dt<\epsilon \int_{f_n >M} g(f_n(t))dt \leq \epsilon$. Now apply DCT to $\int_{f_n \leq M} f_n(t)dt$.

I have assumed that $g>0$. If $g$ has zeros then we can replace $g$ by $\frac {1+g} 2$. This function satisfies the hypothesis and it is strictly positive.