Analysis Problem of the Day 11

Today’s problem appears as Problem 4 on UCLA’s Spring 2011 Analysis Qual:

Problem 4: Let $f_n:[0,1] \to [0,\infty)$ be Borel functions with

$\sup_n \int_0^1 f_n(x)\log(2+f_n(x))dx < \infty.$

Show that if $f_n \to f$ almost everywhere, then $f \in L^1$ and $f_n \to f$ in $L^1.$

Solution: The most powerful convergence result in measure theory is known as the Vitali convergence theorem, which states that if $f_n \in L^p$ is a tight sequence of functions such that $f_n \to f$ a.e. and $\{f_n\}$ is uniformly integrable, then $f_n \to f$ in $L^p$ for all $p<\infty$ (if one works over a finite measure space, the tightness condition may be dropped). In this context, uniform integrability of a subset $S \subset L^p$ has several definitions – the most natural one is that for all $\epsilon>0,$ there exists a $\delta>0$ such that for all $f \in S$ and all measurable sets $A,$ $\|f\|_{L^p(A)}<\epsilon$ whenever $\mu(A)<\delta.$ On finite measure spaces, one can show that this definition is equivalent to the condition that for all $\epsilon>0,$ there exists $N>0$ such that $\sup_{f \in S} \int_{|f|>N} |f|^p dx < \epsilon.$

In particular, since the $f_n$ are nonnegative functions, the statement of the problem implies that if $f_n>N,$ then for some $C>0,$

$C>\int_{f_n>N} f_n(x) \log(2+f_n(x))dx \geq \int_{f_n>N} f_n(x) \log(2+N) dx,$

i.e. $\sup_n \int_{f_n>N} f_n(x)dx < \frac{C}{\log(2+N)}$ for all $n.$ From this one verifies that $f_n \in L^1$ for all $n.$ Also, for $\epsilon>0,$ we see that it suffices to set $N = e^{C\epsilon^{-1}}-2,$ so by the Vitali convergence theorem for $L^1,$ we have that $f \in L^1$ and $f_n \to f$ in $L^1.$

Remark: The proof method shows that any statement of the form

$\sup_n \int_0^1 f_n(x)g(f_n(x))dx < \infty$

for $g:[0,\infty)\to [0,\infty)$ such that $\lim_{x \to \infty} g(x) = \infty$ is enough (along with a.e. convergence) to imply convergence in $L^p$ for $p<\infty.$

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