Failure of Vitali’s Hypotheses in Spike and Oscillation Sequences

Two Different Ways Vitali’s Hypotheses Can Fail

Vitali’s convergence theorem separates two controls on a sequence: convergence in measure describes where the functions differ from their limit, while uniform integrability controls how much integral can be concentrated in large values. Neither condition replaces the other. A spike can occupy a set of vanishing measure and still carry a fixed amount of integral; a bounded oscillatory sequence can be uniformly integrable and still have no limit in measure.

All examples and statements below use the finite, complete measure space $([0,1],\mathcal{L}_{[0,1]},m)$, where $m([0,1])=1$ and $\mathcal{L}_{[0,1]}$ is the Lebesgue sigma-algebra on $[0,1]$. The functions are real-valued and measurable. Convergence in measure to $f$ means that for every $\varepsilon>0$, $m(\{|f_n-f|>\varepsilon\})\to0$. The $L^1$ norm of an integrable function $f$ is $\|f\|_1=\int_{[0,1]}|f|\,dm$; convergence in $L^1$ means $\|f_n-f\|_1\to0$.

Definition: A family $\mathcal{F}$ of integrable real-valued measurable functions on a finite measure space $(X,\mathcal{M},\mu)$ is uniformly integrable if $$ \lim_{K\to\infty}\sup_{f\in\mathcal{F}}\int_{\{|f|>K\}}|f|\,d\mu=0. $$ The same threshold $K$ is used for every member of the family, and the tail set is defined by the strict inequality $|f|>K$. On a finite measure space this condition also implies a uniform bound on the integrals: for any fixed $K$, $\int |f|\,d\mu\le K\mu(X)+\int_{\{|f|>K\}}|f|\,d\mu$.

On a finite measure space, Vitali’s convergence theorem, established in the earlier material on uniform integrability, says that a uniformly integrable sequence which converges in measure to an integrable function converges to that function in $L^1$. The examples below show why both requirements matter. They do not challenge the theorem: each counterexample removes one of its hypotheses.

Small Supports Can Carry Persistent Integral

Worked Example: A Spike Sequence Converging in Measure but Not in L1

For each positive integer $n$, define $f_n(x)=n\chi_{(0,1/n)}(x)$ on $[0,1]$. Each $f_n$ is a nonnegative measurable, integrable function. We first check convergence in measure to the zero function. Fix $\varepsilon>0$. If $n\le\varepsilon$, then $f_n(x)\le\varepsilon$ everywhere, so the strict tail set $\{f_n>\varepsilon\}$ is empty. If $n>\varepsilon$, then $f_n>\varepsilon$ exactly on $(0,1/n)$. Therefore $$ m(\{f_n>\varepsilon\})= \begin{cases} 0,&n\le\varepsilon,\\ 1/n,&n>\varepsilon. \end{cases} $$ As $n\to\infty$, this measure tends to zero for every fixed $\varepsilon>0$. Thus $f_n\to0$ in measure.

But the integral does not tend to zero. The interval $(0,1/n)$ has measure $1/n$, and $f_n$ equals $n$ there, so $$ \|f_n\|_1=\int_{[0,1]} n\chi_{(0,1/n)}\,dm=n\cdot\frac1n=1. $$ Hence $\|f_n-0\|_1=1$ for every $n$, not a sequence tending to zero. Check: substituting $n=4$, the function has height $4$ on an interval of measure $1/4$, and its integral is $4\cdot(1/4)=1$, as the formula predicts.

Now test uniform integrability. Fix any finite $K\ge0$. When $n\le K$, the strict set $\{|f_n|>K\}$ is empty. When $n>K$, that set is $(0,1/n)$, and its tail integral is $$ \int_{\{|f_n|>K\}}|f_n|\,dm=n\cdot\frac1n=1. $$ There is an integer $n>K$, so the supremum over all $n$ is exactly $1$ for every finite $K$. It does not tend to zero as $K\to\infty$. Thus the sequence is not uniformly integrable.

The key distinction is between the measure of the support and the integral over that support. Here the support shrinks at rate $1/n$, while the height grows at rate $n$. Their product remains $1$. Convergence in measure detects the shrinking support, but it does not by itself prevent integral mass from concentrating there.

Proposition: For the spike sequence $f_n=n\chi_{(0,1/n)}$ on $([0,1],\mathcal{L}_{[0,1]},m)$, the sequence converges in measure to zero, does not converge to zero in $L^1$, and is not uniformly integrable.

Proof. For each fixed $\varepsilon>0$, the measure of $\{f_n>\varepsilon\}$ is zero when $n\le\varepsilon$ and $1/n$ when $n>\varepsilon$, which tends to zero. Thus the convergence in measure holds by definition. Since $\int |f_n|\,dm=n\,m((0,1/n))=1$ for every $n$, the $L^1$ distance to zero is always $1$. Finally, for each finite $K\ge0$, choosing any integer $n>K$ gives $\int_{\{|f_n|>K\}}|f_n|\,dm=1$; all such tail integrals are at most $1$, so their supremum is $1$. This tail supremum does not vanish as $K\to\infty$. $\square$

Uniform Integrability Does Not Force Convergence

Uniform integrability is a bound on the tails across the family. It places no requirement that different functions become close to one another. The next example makes the distinction spatial: the functions oscillate more rapidly, while remaining bounded by one.

Worked Example: Bounded Dyadic Oscillations

For $n\ge1$, define $r_n(x)=(-1)^{\lfloor 2^n x\rfloor}$ on $[0,1)$, and set $r_n(1)=0$. These are real-valued measurable functions. On $[0,1)$, each takes only the values $1$ and $-1$. For every $K\ge1$, the strict set $\{|r_n|>K\}$ is empty for all $n$, including at $x=1$, where $r_n(1)=0$. Consequently every tail integral at such a threshold is zero, and the family $\{r_n:n\ge1\}$ is uniformly integrable.

We show that the sequence is not Cauchy in measure. Take integers $m>n$. Partition $[0,1)$ into $2^m$ half-open intervals of length $2^{-m}$, indexed by $k=0,\ldots,2^m-1$. On the interval indexed by $k$, $r_m=(-1)^k$. The intervals split into blocks of $2^{m-n}$ consecutive intervals, each block corresponding to one interval on which $r_n$ is constant. Since $m-n\ge1$, each block has an even number of intervals. Within each block, the values $(-1)^k$ alternate, so exactly half of its intervals have $r_m$ different from the constant value of $r_n$. Therefore the total measure on which $r_m\ne r_n$ is exactly half of $[0,1)$, or $1/2$. On that set $|r_m-r_n|=2$; elsewhere it is zero. In particular, $$ m(\{|r_m-r_n|>1\})=\frac12\qquad(m>n). $$ This prevents the sequence from being Cauchy in measure.

For completeness, convergence in measure implies the Cauchy condition in measure. If $h_j\to h$ in measure and $\varepsilon>0$, the triangle inequality gives $$ \{|h_j-h_k|>\varepsilon\} \subseteq \{|h_j-h|>\varepsilon/2\}\cup\{|h_k-h|>\varepsilon/2\}. $$ Taking measures bounds the left side by the sum of the two measures on the right. Both can be made arbitrarily small by taking $j,k$ sufficiently large, so the sequence is Cauchy in measure. Since the $r_n$ fail this necessary condition, they do not converge in measure. Thus this uniformly integrable sequence cannot converge in $L^1$ to any function: $L^1$ convergence would imply convergence in measure, by the inequality $m(\{|h_j-h|>\varepsilon\})\le\|h_j-h\|_1/\varepsilon$.

Check: the calculation uses the strict event $>1$. A difference of $2$ is counted and a difference of $0$ is not, giving measure $1/2$ for every pair $m>n$. The value assigned at $x=1$ does not change any of these measures because a singleton has Lebesgue measure zero.

Proof. To verify the stated implication from $L^1$ convergence to convergence in measure, let $h_j,h$ be integrable real-valued measurable functions on a measure space and fix $\varepsilon>0$. On the measurable set $A_j=\{|h_j-h|>\varepsilon\}$, the pointwise inequality $|h_j-h|\ge\varepsilon$ holds. Hence $$ \varepsilon\,\mu(A_j)\le\int_{A_j}|h_j-h|\,d\mu\le\|h_j-h\|_1. $$ If the last quantity tends to zero, then $\mu(A_j)\le\|h_j-h\|_1/\varepsilon\to0$. This is convergence in measure by its definition. $\square$

A Control Example: Shrinking Spikes with Shrinking Mass

Worked Example: A Spike Sequence That Does Converge in L1

Define $g_n(x)=n\chi_{(0,1/n^2)}(x)$ on the same finite Lebesgue measure space. The height still grows with $n$, but the support now shrinks faster. For fixed $\varepsilon>0$, the strict set $\{g_n>\varepsilon\}$ is empty when $n\le\varepsilon$, and equals $(0,1/n^2)$ when $n>\varepsilon$. Its measure is therefore $0$ or $1/n^2$, respectively, and tends to zero. Thus $g_n\to0$ in measure. Its $L^1$ norm is $$ \|g_n\|_1=n\,m((0,1/n^2))=n\cdot\frac1{n^2}=\frac1n, $$ which tends to zero; hence $g_n\to0$ in $L^1$.

We also check uniform integrability rather than infer it from convergence. Fix $K\ge0$. If $n\le K$, the strict tail set $\{|g_n|>K\}$ is empty. If $n>K$, it is $(0,1/n^2)$, and its tail integral is $n/n^2=1/n$. Thus $$ \sup_{n\ge1}\int_{\{|g_n|>K\}}|g_n|\,dm =\frac{1}{\lfloor K\rfloor+1}. $$ Indeed, among positive integers strictly greater than $K$, the smallest is $\lfloor K\rfloor+1$, and $1/n$ decreases as $n$ increases. For example, at $K=2$, the first nonempty tail occurs at $n=3$ and has integral $3\cdot(1/9)=1/3$, while later tails are no larger. The displayed supremum tends to zero as $K\to\infty$, as required.

Check: at $n=4$, the support has measure $1/16$, the height is $4$, and $\|g_4\|_1=4\cdot(1/16)=1/4$, agreeing with $1/n$. At threshold $K=2$, its tail integral is $1/4$, which is at most the family supremum $1/3$.

Reading the Two Hypotheses Correctly

The two failures are logically different. In the first example, convergence in measure holds but uniform integrability fails; the integral remains trapped in increasingly large values on increasingly small sets. In the second, uniform integrability holds but convergence in measure fails; controlling large values cannot make the functions settle down. The control example has both properties, and its $L^1$ norm tends to zero, in agreement with Vitali’s theorem.

Sequence on $[0,1]$	Converges in measure to zero?	Uniformly integrable?	$L^1$ behavior
$n\chi_{(0,1/n)}$	Yes	No	Norm is $1$ for every $n$
$r_n$, the dyadic oscillations	No	Yes	Cannot converge in $L^1$
$n\chi_{(0,1/n^2)}$	Yes	Yes	Norm is $1/n$, tending to zero

A common mistake is to argue that the first spike sequence has integral tending to zero because the measure of its support tends to zero. That calculation omits the height. The incorrect computation would be “the support has measure $1/n$, so the integral tends to $0$.” The correction is to multiply the function’s value by the support measure: $\int f_n\,dm=n\cdot(1/n)=1$. Small support alone is not enough; uniform integrability is the condition that rules out persistent tail mass.

A second pitfall is to regard uniform integrability as a convergence condition. For the oscillatory sequence, all values have absolute value at most one, so every tail above $K=1$ is empty. That proves uniform integrability, not convergence. The fixed positive measure of the sets $\{|r_m-r_n|>1\}$ proves the separate failure of convergence in measure.

Check Your Understanding

Use the definitions, including the strict tail-set inequalities, to justify each answer.

For $u_n=n\chi_{(0,1/n)}$, compute $m(\{u_n>\varepsilon\})$ for fixed $\varepsilon>0$ and compute $\|u_n\|_1$.
For the same $u_n$, fix $K\ge0$. What is $\sup_n\int_{\{|u_n|>K\}}|u_n|\,dm$, and why does this rule out uniform integrability?
For $r_n(x)=(-1)^{\lfloor2^nx\rfloor}$ on $[0,1)$, explain why $m(\{|r_m-r_n|>1\})=1/2$ whenever $m>n$.
For $g_n=n\chi_{(0,1/n^2)}$, compute the tail-integral supremum at $K=3$, stating which integer first has a nonempty strict tail.
Why does $L^1$ convergence imply convergence in measure, and which inequality gives the implication?

Sequence on \([0,1]\)	Converges in measure to zero?	Uniformly integrable?	\(L^1\) behavior
\(n\chi_{(0,1/n)}\)	Yes	No	Norm is \(1\) for every \(n\)
\(r_n\), the dyadic oscillations	No	Yes	Cannot converge in \(L^1\)
\(n\chi_{(0,1/n^2)}\)	Yes	Yes	Norm is \(1/n\), tending to zero