A Weighting Argument for Pointwise Convergence to Convergence in Measure

The mechanism

Change the scale at each point

Pointwise information need not give uniform control. This argument introduces a positive measurable weight that adjusts the scale from point to point. The weighted sequence can then be controlled using a decomposition argument or dominated convergence.

What becomes elementary: Choosing the weight turns local convergence information into a statement about the measure of exceptional sets.

For the shortest proof of the title’s conclusion, use the dominated-convergence argument in §4: it gives convergence in L¹ and hence in measure. The construction in §§2–3 does more work to obtain uniform convergence on a conull set. In each route, track positivity and measurability of the weight.

Explore further constructions. Edition dates, sources and reference chapters identify the scope of each study.

Overview

Let $(f_n)$ be a sequence of real-valued Lebesgue measurable functions on $\mathbb{R}$ such that $f_n(x)\to 0$ for every $x\in\mathbb{R}$ . We give a detailed proof that there exists a strictly positive measurable function $h$ such that $f_nh\to 0$ in measure. The main proof is an expanded version of a Math StackExchange answer of the author: it uses a finite-measure exhaustion, Egorov’s theorem on each finite-measure piece, and a pointwise-boundedness stratification. The construction actually proves the stronger conclusion that $f_nh\to 0$ uniformly on a conull measurable subset of $\mathbb{R}$ . We also include a shorter dominated-convergence proof of the original statement and record the corresponding sigma-finite version.

1 · Introduction

On a finite measure space, almost everywhere convergence implies convergence in measure. On an infinite measure space, such as $(\mathbb{R},m)$ with Lebesgue measure, this implication is false in general. The problem considered here asks whether one can multiply by a positive measurable weight in order to force convergence in measure.

The answer is yes. More precisely, if $f_n:\mathbb{R}\to\mathbb{R}$ are measurable and $f_n(x)\to 0$ pointwise, then there is a strictly positive measurable function $h:\mathbb{R}\to(0,\infty)$ such that

f_nh\longrightarrow 0

in measure. The construction below proves something stronger:

h

can be chosen so that

f_nh\to 0

uniformly on a measurable set whose complement has Lebesgue measure zero.

The key idea is to combine two elementary consequences of pointwise convergence. First, after decomposing $\mathbb{R}$ into countably many finite-measure sets, Egorov’s theorem gives countably many large pieces on which the convergence is uniform. Second, at each fixed point $x$ , the scalar sequence $(f_n(x))_n$ is bounded, so $\mathbb{R}$ is the union of the level sets on which the pointwise envelope $\sup_n |f_n|$ is bounded. The weight $h$ is then made small on the intersections of these two decompositions.

Throughout this note, convergence in measure on $\mathbb{R}$ means the global condition

m\bigl(\{x\in\mathbb{R}: |g_n(x)|>\varepsilon\}\bigr)\longrightarrow 0 \qquad\text{for every }\varepsilon>0.

Equivalently, one may use

\geq \varepsilon

instead of

>\varepsilon

. All indices below are positive integers, so

\mathbb{N}=\{1,2,\ldots\}

. For suprema of nonnegative quantities over an empty set, we use the value

0

2 · The strengthened result on $\mathbb{R}$

We first state the stronger form proved by the weighting construction.

Theorem 2.1 (Conull uniformization by a positive weight). Let $f_n:\mathbb{R}\to\mathbb{R}$ be Lebesgue measurable functions such that
$f_n(x)\to 0 \qquad\text{for every }x\in\mathbb{R}.$ Then there exist a measurable set $A\subseteq\mathbb{R}$ and a strictly positive finite-valued measurable function $h:\mathbb{R}\to(0,\infty)$ such that $m(\mathbb{R}\setminus A)=0$ and $\sup_{x\in A}|f_n(x)h(x)|\longrightarrow 0.$ In particular, $f_nh\to 0$ in measure on $\mathbb{R}$ .

The proof is split into two organizing lemmas.

Lemma 2.2 (Countably many uniform-convergence pieces). There exist measurable sets $A_1,A_2,\dots\subseteq\mathbb{R}$ such that
$m\left(\mathbb{R}\setminus \bigcup_{i=1}^\infty A_i\right)=0,$ and such that $f_n\to 0$ uniformly on each $A_i$ .

Proof. Choose a finite-measure exhaustion of $\mathbb{R}$ ; for example, take

E_r=[-r,r],\qquad r\geq 1.

For each pair

r,k\geq 1

, Egorov’s theorem applied on

E_r

gives a measurable set

A_{r,k}\subseteq E_r

such that

m(E_r\setminus A_{r,k})<\frac{1}{k}

and

f_n\to 0

uniformly on

A_{r,k}

For fixed $r$ , we have

E_r\setminus \bigcup_{k=1}^\infty A_{r,k} =\bigcap_{k=1}^\infty (E_r\setminus A_{r,k}),

and hence, for every

k

m\left(E_r\setminus \bigcup_{k=1}^\infty A_{r,k}\right) \leq m(E_r\setminus A_{r,k})<\frac{1}{k}.

Therefore

m\left(E_r\setminus \bigcup_{k=1}^\infty A_{r,k}\right)=0.

Let

A:=\bigcup_{r,k\geq 1}A_{r,k}.

Since

\mathbb{R}=\bigcup_r E_r

, it follows that

\mathbb{R}\setminus A \subseteq \bigcup_{r=1}^\infty \left(E_r\setminus \bigcup_{k=1}^\infty A_{r,k}\right),

m(\mathbb{R}\setminus A)=0

. Finally, reindex the countable family

\{A_{r,k}:r,k\geq 1\}

A_1,A_2,\dots

. Each reindexed set is measurable, and

f_n\to 0

uniformly on each one. ◻

Lemma 2.3 (Pointwise boundedness stratification). Define
$M(x):=\sup_{n\geq 1}|f_n(x)|.$ Then $M$ is measurable and finite everywhere. Consequently, the sets $B_j:=\{x\in\mathbb{R}:M(x)\leq j\},\qquad j\geq 1,$ are measurable and satisfy $\mathbb{R}=\bigcup_{j=1}^\infty B_j.$ Moreover, on $B_j$ one has $|f_n(x)|\leq j \qquad\text{for every }n\geq 1.$

Proof. Since each $f_n$ is measurable, so is $|f_n|$ , and therefore the countable supremum

M=\sup_{n\geq 1}|f_n|

is measurable. For each fixed

x\in\mathbb{R}

, the numerical sequence

(f_n(x))_{n\geq 1}

converges to

0

, hence is bounded. Thus

M(x)<\infty

for every

x

. The remaining assertions follow immediately from the definition of

B_j

. ◻

Proof of Theorem 2.1. Let $A=\bigcup_i A_i$ , where the sets $A_i$ come from Lemma 2.2. Let $B_j$ be the measurable sets from Lemma 2.3. Define

h(x) := \sum_{i,j\geq 1}\frac{1}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) +\mathbf 1_{\mathbb{R}\setminus A}(x).

This is a nonnegative measurable function, being a countable sum of nonnegative measurable functions.

We first check that $h$ is finite-valued and strictly positive. Since

\sum_{i,j\geq 1}\frac{1}{j2^{ij}} \leq \sum_{i,j\geq 1}\frac{1}{2^{ij}} = \sum_{i=1}^\infty \frac{1}{2^i-1} \leq \sum_{i=1}^\infty \frac{1}{2^{i-1}} =2,

the defining series is bounded above by

2

everywhere. Hence

h

is finite. If

x\in\mathbb{R}\setminus A

, then

h(x)\geq 1

. If

x\in A

, then

x\in A_i

for some

i

, and since

\mathbb{R}=\bigcup_j B_j

, also

x\in B_j

for some

j

. The corresponding summand is then positive. Thus

h(x)>0

for every

x\in\mathbb{R}

It remains to prove uniform convergence on $A$ . Fix $\varepsilon>0$ . Because

\sum_{i,j\geq 1}2^{-ij}<\infty,

we may choose

s\geq 1

so large that

\sum_{\max\{i,j\}>s}2^{-ij}<\frac{\varepsilon}{2}.

For

x\in A

, the term

\mathbf 1_{\mathbb{R}\setminus A}(x)

vanishes, and therefore

\begin{aligned} |f_n(x)h(x)| &\leq \sum_{i,j\geq 1} \frac{|f_n(x)|}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) \\ &= \sum_{1\leq i,j\leq s} \frac{|f_n(x)|}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) + \sum_{\max\{i,j\}>s} \frac{|f_n(x)|}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x). \end{aligned}

A_i\cap B_j

, Lemma 2.3 gives

|f_n(x)|\leq j

. Hence the tail is bounded uniformly in

x\in A

and

n

\sum_{\max\{i,j\}>s}2^{-ij}<\frac{\varepsilon}{2}.

For the finite block, we have

\begin{aligned} \sup_{x\in A} \sum_{1\leq i,j\leq s} \frac{|f_n(x)|}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) &\leq \sum_{1\leq i,j\leq s} \frac{1}{j2^{ij}}\sup_{x\in A_i}|f_n(x)|. \end{aligned}

For each fixed

i

f_n\to 0

uniformly on

A_i

, and the sum above has only finitely many indices. Therefore the finite block tends to

0

uniformly on

A

. Thus there exists

N

such that for all

n\geq N

\sup_{x\in A} \sum_{1\leq i,j\leq s} \frac{|f_n(x)|}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) <\frac{\varepsilon}{2}.

Combining the finite-block estimate with the tail estimate gives

|f_n(x)h(x)|<\varepsilon \qquad\text{for all }x\in A\text{ and all }n\geq N.

Hence

\sup_{x\in A}|f_n(x)h(x)|\to 0.

Finally, since $m(\mathbb{R}\setminus A)=0$ , uniform convergence on $A$ implies convergence in measure. Indeed, given $\varepsilon>0$ , for all sufficiently large $n$ ,

\{x\in\mathbb{R}:|f_n(x)h(x)|>\varepsilon\} \subseteq \mathbb{R}\setminus A,

and the right-hand side has measure zero. Therefore

f_nh\to 0

in measure. ◻

3 · A construction with disjoint pieces

We can replace the double series by first making the intersections of the Egorov pieces $A_i$ and the boundedness pieces $B_j$ disjoint.

Enumerate all pairs $(i,j)\in\mathbb{N}^2$ as

(i_1,j_1),(i_2,j_2),(i_3,j_3),\dots,

and set

C_p:=A_{i_p}\cap B_{j_p}.

Now disjointify these sets by defining

D_1=C_1, \qquad D_p=C_p\setminus \bigcup_{q<p}C_q \quad(p\geq 2).

Then the

D_p

are measurable, pairwise disjoint, and

A=\bigcup_{p=1}^\infty D_p.

Moreover, if

x\in D_p

, then

x\in A_{i_p}

and

x\in B_{j_p}

, so

f_n\to 0

uniformly on

D_p

and

|f_n(x)|\leq j_p \qquad (x\in D_p,\ n\geq 1).

Define instead

\widetilde h(x) := \sum_{p=1}^\infty \frac{1}{p j_p}\mathbf 1_{D_p}(x) +\mathbf 1_{\mathbb{R}\setminus A}(x).

Because the

D_p

are disjoint, this simply says that

\widetilde h(x)=\frac{1}{p j_p}\quad\text{if }x\in D_p, \qquad \widetilde h(x)=1\quad\text{if }x\notin A.

Thus

\widetilde h

is measurable, finite-valued, and strictly positive.

To prove uniform convergence, fix $\varepsilon>0$ and choose $P$ so large that $1/P<\varepsilon$ . If $p>P$ and $x\in D_p$ , then for every $n$ ,

|f_n(x)\widetilde h(x)| \leq j_p\cdot \frac{1}{p j_p} =\frac{1}{p} <\varepsilon.

On the finite union

D_1\cup\cdots\cup D_P

, uniform convergence follows from uniform convergence on the corresponding finitely many sets

A_{i_p}

. More explicitly, for each

p\leq P

, choose

N_p

so that

\sup_{x\in D_p}|f_n(x)|<\varepsilon p j_p \qquad(n\geq N_p).

With

N=\max_{1\leq p\leq P}N_p

, we get for every

n\geq N

and every

x\in D_p

p\leq P

|f_n(x)\widetilde h(x)| <\varepsilon.

Together with the tail estimate for

p>P

, this proves

\sup_{x\in A}|f_n(x)\widetilde h(x)|\leq \varepsilon

for all sufficiently large

n

. Hence

f_n\widetilde h\to 0

uniformly on

A

, and therefore in measure.

Remark 3.1. This disjointified proof is the same idea as the double-series proof. The sets $A_i$ encode where convergence is already uniform; the sets $B_j$ encode where the whole sequence is uniformly bounded; and the coefficients of the weight shrink sufficiently fast as one moves through the countable list of pieces.

4 · A shorter proof of the original convergence-in-measure statement

A different choice of weight gives convergence in $L^1$ , and hence in measure, by dominated convergence.

Let

M(x):=\sup_{n\geq 1}|f_n(x)|.

As above,

M

is measurable and finite everywhere. Set

g(x):=e^{-|x|}

and define

h_0(x):=\frac{g(x)}{1+M(x)}.

Then

h_0

is measurable and strictly positive. Moreover,

|f_n(x)h_0(x)| =|f_n(x)|\frac{g(x)}{1+M(x)} \leq g(x).

Since

g\in L^1(\mathbb{R})

and

f_n(x)h_0(x)\to 0

pointwise, the dominated convergence theorem gives

\int_{\mathbb{R}}|f_nh_0|\,dm\longrightarrow 0.

Thus

f_nh_0\to 0

L^1(\mathbb{R})

, and hence in measure, since Markov’s inequality gives

m\bigl(\{x\in\mathbb{R}:|f_n(x)h_0(x)|>\varepsilon\}\bigr) \leq \frac{1}{\varepsilon}\int_{\mathbb{R}}|f_nh_0|\,dm \longrightarrow 0

for every

\varepsilon>0

Remark 4.1. The two weights give different conclusions: the dominated-convergence weight gives $L^1$ convergence, while the Egorov construction gives uniform convergence on a conull set.

5 · The sigma-finite form

Nothing in the Egorov argument is special to the topology of $\mathbb{R}$ . The relevant measure-theoretic hypothesis is sigma-finiteness.

Theorem 5.1 (Sigma-finite version). Let $(X,\mathcal{A},\mu)$ be a sigma-finite measure space, and let $f_n:X\to\mathbb{R}$ be measurable functions such that $f_n(x)\to 0$ for $\mu$ -almost every $x\in X$ . Then there exist a measurable set $A\subseteq X$ with $\mu(X\setminus A)=0$ and a strictly positive finite-valued measurable function $h:X\to(0,\infty)$ such that
$\sup_{x\in A}|f_n(x)h(x)|\longrightarrow 0.$ Consequently $f_nh\to 0$ in measure.

Proof. Let

X_0:=\{x\in X:f_n(x)\to 0\}.

This set is measurable, since

X_0= \bigcap_{\ell=1}^\infty \bigcup_{N=1}^\infty \bigcap_{n\geq N}\{x\in X:|f_n(x)|<1/\ell\},

and it is conull by hypothesis. Choose a countable cover

X=\bigcup_{r=1}^\infty E_r

with

\mu(E_r)<\infty

. Applying Egorov’s theorem on each finite-measure set

E_r\cap X_0

, exactly as in Lemma 2.2, gives countably many measurable sets

A_1,A_2,\dots\subseteq X_0

such that

\mu\left(X\setminus \bigcup_{i=1}^\infty A_i\right)=0,

and

f_n\to 0

uniformly on each

A_i

. Put

A:=\bigcup_{i=1}^\infty A_i.

On $X$ , define the extended-valued measurable envelope

M(x):=\sup_{n\geq 1}|f_n(x)|.

It is finite on

X_0

. For

j\geq 1

, define

B_j:=X_0\cap\{x\in X:M(x)\leq j\}.

Then the

B_j

are measurable,

X_0=\bigcup_j B_j

, and

|f_n|\leq j

B_j

for all

n

Now define

h(x) := \sum_{i,j\geq 1}\frac{1}{j2^{ij}}\mathbf 1_{A_i\cap B_j}(x) +\mathbf 1_{X\setminus A}(x).

The same estimates as in the proof of Theorem 2.1 show that

h

is measurable, finite-valued, strictly positive everywhere, and that

f_nh\to 0

uniformly on

A

. Since

\mu(X\setminus A)=0

, convergence in measure follows. ◻

The shorter $L^1$ proof also has a sigma-finite version.

Proposition 5.2 (Sigma-finite dominated-convergence proof). Let $(X,\mathcal{A},\mu)$ be sigma-finite, and suppose $f_n:X\to\mathbb{R}$ are measurable and $f_n\to 0$ almost everywhere. Then there is a strictly positive finite-valued measurable function $h$ such that
$\int_X |f_nh|\,d\mu\longrightarrow 0.$ In particular, $f_nh\to 0$ in measure.

Proof. Let $X_0=\{x:f_n(x)\to 0\}$ , and let

M(x):=\sup_{n\geq 1}|f_n(x)|.

The function

M

is an extended-valued measurable function and is finite on

X_0

. Define the finite-valued measurable function

\widetilde M(x):= \begin{cases} M(x), & x\in X_0,\\ 0, & x\notin X_0. \end{cases}

Since $\mu$ is sigma-finite, choose measurable sets $E_r$ with $X=\bigcup_r E_r$ and $\mu(E_r)<\infty$ . Passing to the disjoint refinement

F_1=E_1, \qquad F_r=E_r\setminus\bigcup_{q<r}E_q\quad(r\geq 2),

we obtain a measurable partition of

X

with

\mu(F_r)<\infty

. Define

g(x):=\sum_{r=1}^\infty \frac{2^{-r}}{1+\mu(F_r)}\mathbf 1_{F_r}(x).

Then

g

is strictly positive, finite-valued, measurable, and integrable, since

\int_X g\,d\mu = \sum_{r=1}^\infty \frac{2^{-r}\mu(F_r)}{1+\mu(F_r)} \leq \sum_{r=1}^\infty 2^{-r}<\infty.

Define

h(x):=\frac{g(x)}{1+\widetilde M(x)}.

Then

h

is strictly positive, finite-valued, and measurable. On the conull set

X_0

|f_nh| \leq g \qquad\text{and}\qquad f_nh\to 0.

The dominated convergence theorem therefore gives

\int_X |f_nh|\,d\mu\to 0.

The convergence in measure follows from Markov’s inequality. ◻

6 · Concluding remarks

The Egorov construction combines two countable stratifications:

pieces on which convergence is uniform, supplied here by Egorov’s theorem on finite-measure pieces;
pieces on which the whole sequence is uniformly bounded, supplied here by the pointwise envelope $M=\sup_n |f_n|$ .

The weight is then chosen small enough on the intersections of the two stratifications. The role of sigma-finiteness is precisely to make the first stratification countable through finite-measure exhaustion. The role of pointwise convergence is not only to give the limiting value $0$ , but also to guarantee pointwise boundedness of the sequence, which is what makes the second stratification possible.

Henry Shin, Answer to: Let $f_n$ be real and measurable with $f_n(x)\to 0$ ; show there is a positive measurable function $h$ such that $f_nh\to 0$ in measure, Mathematics Stack Exchange, answered July 1, 2021. https://math.stackexchange.com/questions/4186426/let-f-n-be-real-measurable-w-f-nx-rightarrow-0-show-there-is-a-positiv

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley, 1999.

H. L. Royden and P. M. Fitzpatrick, Real Analysis, 4th ed., Pearson, 2010.