Analysis Problem of the Day 74

Today’s problem appeared as Problem 3 on the UCLA Spring 2012 Analysis Qual:

Problem 3. Let $f \in L^1([0,1]),$ and define

$f_n(x) = n\int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x)dx, \quad x \in \left[\frac{k}{n},\frac{k+1}{n}\right), \quad k=0,...,n-1.$

Show that $f_n \to f$ a.e. and in $L^1.$

Solution: Notice that $f_n$ is defined at each point as its average on one of $n$ equally spaced intervals that contains the point. In particular, one of the tools we have is the general version of the Lebesgue differentiation theorem, which says that the averages of a locally integrable function over sets shrinking nicely to a point converge to the value of the function at that point a.e. This immediately yields that $f_n \to f$ a.e.

To show $f_n \to f$ in $L^1,$ we use the notion of uniform integrability, as a.e. convergence of a uniformly integrable sequence of functions implies convergence in $L^1.$ It suffices to show that for all $\epsilon>0$ there exists a $\delta>0$ such that $\mu(A)<\delta$ implies $\sup_n \int_A |f_n| dx < \epsilon.$ Indeed, note that since $f \in L^1,$ for any $\epsilon >0$ there exists $\delta>0$ such that $\int_B |f|dx<\epsilon$ whenever $\mu(B)<\delta.$ Now, for each subinterval $I_0,...,I_{n-1}$ used in the construction of $f_n,$ define $A_k:=I_k \cap A,$ and note that there exists $B_k \subseteq I_k$ such that $\mu(B_k) \leq \mu(A_k)$ and $\int_{A_k} |f_n| dx \leq \int_{B_k} |f|dx.$ This can be show by using the layer-cake decomposition of $f_n$ on $I_k$ by letting $B_k=\{x: |f|>\lambda_k\},$ where

$\lambda_k := \sup \left\{j>0: \int_{B_j} |f| dx \geq \int_{A_k} |f_n|dx\right\}.$

Such a $\lambda_k$ exists since

$\int_{B_0} |f|dx =\int_{I_k} |f| dx \geq \int_{I_k} |f_n| dx \geq \int_{A_k} |f_n| dx$

by triangle inequality, and $\mu(B_k) \leq \mu(A_k),$ since otherwise, for some $\epsilon>0$ and $\delta>0$ small enough such that $\int_{\lambda_k \leq |f| \leq \lambda_k +\delta}|f|dx< \epsilon n \int_{I_k} |f|dx,$ one would have

$\int_{A_k} |f_n|dx = \mu(A_k) n \int_{I_k} |f| dx \leq (\mu(B_k) - \epsilon) n \int_{I_k} |f|dx \leq \int_{|f|>\lambda_k + \delta} |f| dx,$

which directly contradicts the definition of $\lambda_k.$ It follows that

$\int_A |f_n|dx = \sum_{k=0}^{n-1} \int_{A_k} |f_n|dx \leq \sum_{k=0}^{n-1} \int_{B_k} |f|dx = \int_B |f|dx<\epsilon$

for $B:=\bigcup_{k=0}^{n-1} B_k$ and $\mu(B) \leq \mu(A)<\delta.$ Thus, $\{f_n\}$ is uniformly integrable and converges a.e. to $f$ on a finite measure space, so $f_n \to f$ in $L^1$ by the Vitali convergence theorem.

Remark: The techniques utilized in this problem are similar to those of decreasing rearrangements – in other words, one takes the mass of an integrable function at large values and shows that it dominates the mass of the function over any other set of the same measure.

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