exercise on uniform integrability

Question

I cannot figure out the following exercise:

Let $F$ be the family of functions $f$ on $[0,1]$, each of which is (Lebesgue) integrable over $[0,1]$ and has $\int_a^b|f|\le b-a$ for all $[a,b]\subseteq[0,1]$. Is $F$ uniformly integrable over $[0,1]$?

Answer 1

The answer is 'yes'.

To prove this, we first show, as suggested by Did, that for any $f$ in $F$ one has $|f|\le 1$ a.e. on $[0,1]$. For that, denote $A_n=\{x\in[0,1] : |f(x)|>1+\frac{1}{n} \}$, and $A=\{x\in [0,1]:|f(x)|>1\}$. Obviously, $A_n\subseteq A_{n+1}$ for all $n$, and $m(A)=\lim_{n\to\infty}m(A_n)$. We prove that $m(A_n)=0$ for every value of $n$, which we fix from now on. On one hand, $$ \int_{A_n}|f|\ge m(A_n)\cdot(1+\tfrac1n) $$ by Chebyshev's inequality. On the other hand, for any $\epsilon>0$ one can find an open cover $U_\epsilon\supseteq A_n$ such that $m(A_n)\le m(U_\epsilon)\le m(A_n)+\epsilon$. Being an open set, $U_\epsilon$ is a disjoint countable union of open intervals: $U_\epsilon=\bigsqcup_{k=1}^\infty I_k$. By the countable additivity of Lebesgue integral, $$\int_{A_n}|f|\le\int_{U_\epsilon}|f|=\sum_{k=1}^n\int_{I_k}|f|\le \sum_{k=1}^n\ell(I_k)=m(U_\epsilon)\le m(A_n)+\epsilon.$$ So we have: $$ m(A_n)\cdot(1+\tfrac1n)\le\int_{A_n}|f|\le m(A_n)+\epsilon. $$ As $n$ is fixed and $\epsilon$ can be made arbitrarily small, we conclude that $m(A_n)=0$, and hence $m(A)=\lim_{n\to\infty}m(A_n)=0$. This proves that $|f|\le 1$ almost everywhere on $[0,1]$.

Now, if $B$ is a measurable subset of $[0,1]$ with $m(B)<\delta$, one has $\int_B|f|\le\int_B1=m(B)<\delta$, so one can set $\delta=\epsilon$ in the definition of the uniform integrability.