Modes of Convergence Egorov and Lusin: Examples and Counterexamples

If $2^k\le n<2^{k+1}$, then $f_n$ is the indicator of a dyadic interval of length $2^{-k}$, so ${\int }_0^1\mathrm{\mid }{f}_n\mathrm{\mid }d\lambda =2^{-k}\mathrm{.}$

Fix $x\in [0,1]$. In each generation $k\ge 1$ there is exactly one interval $I_{k,j_k}$ containing $x$, so $f_{2^k+j_k}(x)=1$. The same generation also contains intervals missing $x$, hence infinitely many indices $n$ with $f_n(x)=0$. So $(f_n(x))$ takes the values $0$ and $1$ infinitely often and therefore does not converge.

Since $2^k\le n<2^{k+1}$ forces $k\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, step 1.1 gives ${\int }_0^1\mathrm{\mid }{f}_n\mathrm{\mid }d\lambda \to 0$. Therefore $f_n\to 0$ in $L^1([0,1])$ by [L2].

Fix $\epsilon \in (0,1)$. When $2^k\le n<2^{k+1}$, the set $\{\mathrm{\mid }{f}_n-0\mathrm{\mid }>\epsilon \}$ is exactly the support interval of $f_n$, so it has measure $2^{-k}$. Thus these bad-set measures tend to $0$, and [L1] gives $f_n\to 0$ in measure.

Step 2.2 proves convergence in measure, step 2.1 proves convergence in $L^1$, and step 1.2 shows that [L3] fails at every point.

By [L1], $g_0=f_1={\chi }_{[0,1]}$, and for $k\ge 1$ the index $2^k$ picks the first dyadic interval of generation $k$, so $g_k={\chi }_{[0,2^{-k})}\mathrm{.}$

If $x\in (0,1]$, choose $K$ with $2^{-K}<x$. Then for every $k\ge K$ one has $x\mathrm{\notin }[0,2^{-k})$, so $g_k(x)=0$. Thus $g_k(x)\to 0$ for every $x\in (0,1]$, while $g_k(0)=1$ for all $k$.

Let $\epsilon >0$, put $\delta :=\min \{\epsilon \mathrm{/}2,1\mathrm{/}2\}$, and take $E:=[0,\delta )$. Then $\lambda (E)=\delta <\epsilon$, and if $x\in [\delta ,1]$ and $2^{-k}<\delta$ then $g_k(x)=0$. So $g_k\to 0$ uniformly on $[\delta ,1]$, which is almost-uniform convergence.

Step 2.1 exhibits an explicit almost-everywhere convergent subsequence of the typewriter family, matching the general existence promised by [L2], and step 2.2 strengthens it to almost-uniform convergence, matching [L3].

Fix $x\in \mathbb{R}$. If $n>x+1$, then $x\mathrm{\notin }[n,n+1]$, so $f_n(x)=0$. Hence $f_n(x)\to 0$ for every $x\in \mathbb{R}$, and therefore almost everywhere by [L1].

For every $n$ one has $\{\mathrm{\mid }{f}_n-0\mathrm{\mid }>1\mathrm{/}2\}=[n,n+1]$, whose Lebesgue measure is $1$. So the bad-set measures do not tend to $0$, and [L2] fails.

This sequence satisfies the premise of the refuted statement and violates its conclusion.

By [L1], the sequence $f_n={\chi }_{[n,n+1]}$ converges almost everywhere to $0$.

If this convergence were almost uniform, then [L2] would force convergence in measure as well. But [L1] says that convergence in measure fails. So the convergence is not almost uniform.

Fix $x\in [0,1]$. If $x>0$, then $1\mathrm{/}n<x$ for all large $n$, so $f_n(x)=0$ eventually; and $f_n(0)=0$ for every $n$. Thus $f_n(x)\to 0$ for every $x\in [0,1]$, hence almost everywhere by [L1].

Fix $\epsilon >0$. For all $n>\epsilon$ one has $\{\mathrm{\mid }{f}_n-0\mathrm{\mid }>\epsilon \}=(0,1\mathrm{/}n)$, whose measure is $1\mathrm{/}n\to 0$. So $f_n\to 0$ in measure by [L2].

For every $n\ge 1$, [given, L3, algebra] ${\int }_0^1\mathrm{\mid }{f}_n\mathrm{\mid }d\lambda ={\int }_0^{1\mathrm{/}n}nd\lambda =1.$ So the $L^1$ errors from $0$ never tend to $0$, and [L3] fails.

The sequence converges almost everywhere and in measure to $0$, but not in $L^1$.

If $0\le x<1$, then $x^k\to 0$, while $1^k=1$ for every $k$. Thus $f_k\to f$ pointwise on $[0,1]$.

Put $E:=(1-\epsilon ,1]$. Then $\lambda (E)=\epsilon$, and for $x\in [0,1-\epsilon ]$ one has $f(x)=0$ and $\mathrm{\mid }{f}_k(x)-f(x)\mathrm{\mid }=x^k\le (1-\epsilon {)}^k$. Since $(1-\epsilon {)}^k\to 0$, the convergence is uniform on $[0,1-\epsilon ]$.

This is the almost-uniform conclusion predicted abstractly by [L1], and here the exceptional set and the uniform estimate are explicit.

By [L1], enumerate the rationals in $[0,1]$ as $(q_j{)}_{j\ge 1}$. For each $j$, choose an open interval $U_j$ centred at $q_j$ with length below $\epsilon 2^{-j-1}$, and put $U:={\bigcup }_{j\ge 1}{U}_j$. Then $\mathbb{Q}\cap [0,1]\subseteq U$, and [L2] gives $\lambda (U)\le {\sum }_{j=1}^{\mathrm{\infty }}\lambda (U_j)<\epsilon \mathrm{.}$

Put $F:=[0,1]\setminus U$. Since $U$ is open, [L4] makes $F$ closed, and $F$ contains no rationals. Therefore $D{\mathrm{\mid }}_F=0$, so the restriction is continuous. Also $[0,1]\setminus F=U\cap [0,1]\subseteq U$, so [L3] and step 1.1 give $\lambda ([0,1]\setminus F)\le \lambda (U)<\epsilon$.

By [L5], every neighbourhood of every point of $[0,1]$ meets both $\mathbb{Q}$ and its complement, so $D$ is nowhere continuous on $[0,1]$. Thus step 2.1 is about the restriction $D{\mathrm{\mid }}_F$, not continuity of $D$ at the points of $F$.

The series ${\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k^2}$ converges to a value below $1$, so the intervals $I_k$ are pairwise disjoint subsets of $[0,1]$. Also $\int \mathrm{\mid }{f}_k\mathrm{\mid }d\lambda =k\lambda (I_k)=\frac{1}{2k}\mathrm{.}$

If $M>0$ and $k\le M$, then $\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}=\mathrm{\varnothing }$; if $k>M$, then $\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}=I_k$ and ${\int }_{\{\mathrm{\mid }{f}_k\mathrm{\mid }>M\}}\mathrm{\mid }{f}_k\mathrm{\mid }d\lambda =\frac{1}{2k}\le \frac{1}{2\lfloor M\rfloor +2}\mathrm{.}$ Hence the family is uniformly integrable by [L1].

If $g$ were an integrable majorant for all $f_k$, then $\mathrm{\mid }g\mathrm{\mid }\ge g\ge k$ almost everywhere on $I_k$. Since the intervals are pairwise disjoint, $\int \mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}{\int }_{I_k}\mathrm{\mid }g\mathrm{\mid }d\lambda \ge {\sum }_{k=1}^{\mathrm{\infty }}\frac{1}{2k}=+\mathrm{\infty },$ contradicting integrability of $g$.

This family is therefore uniformly integrable but has no single integrable majorant. The example is exactly the failure of the false statement paired with [L2].