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A null set has Lebesgue outer measure zero. Equivalently, for every $\epsilon >0$ it can be covered by countably many open intervals whose total length is less than $\epsilon$. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

If $A\subseteq E$ and $E$ is a null set, every interval cover of $E$ also covers $A$. Thus monotonicity of Lebesgue outer measure gives $0\le m^{\ast }(A)\le m^{\ast }(E)=0$. Hence every subset of a null set is null, and in particular is Lebesgue measurable.

If $a\ge 0$, the set $\{f>a\}$ is contained in $E$, since $f=0$ outside $E$. It is therefore a null set and is Lebesgue measurable. If $a<0$, its complement $\{f\le a\}$ is contained in $E$ and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving $f$ is a measurable function.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

Put $G_N={\sum }_{n=1}^N|f_n|$. These are nonnegative measurable functions increasing to $G={\sum }_{n=1}^{\infty }|f_n|$, possibly initially with value infinity. Monotone convergence theorem givesAn infinite value on a set of positive measure would force an infinite integral. Thus $G$ is finite almost everywhere and, after setting it to zero on its null exceptional set if desired, is integrable.

Where $G$ is finite, the real series ${\sum }_n{f}_n$ is absolutely convergent. Define its sum to be $h$ there and zero on the exceptional null set. Measurability follows as an almost-everywhere limit of partial sums, and $|h|\le G$ almost everywhere gives integrability. Every partial sum is dominated by $G$, so dominated convergence theorem yieldsIn fact the convergence is also in the integrable-function norm:

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

Choose measurable versions of the two Lebesgue integrable functions and consider $F(x,y)=g(x-y)h(y)$. Translation and the corresponding linear change of variables preserve Lebesgue measurability; one can first use Borel versions and then disregard the resulting null-set changes. Tonelli's theorem and translation invariance giveFubini theorem therefore shows that the section in $y$ is integrable for almost every $x$. Define $(g\ast h)(x)$ by that section integral where it exists and by zero on the exceptional null set. Fubini theorem also makes this a measurable Lebesgue integrable function, withIts almost-everywhere defining formula is precisely the desired convolution. This construction explains why an everywhere-existing pointwise integral is not required.

Fubini's theorem for the product of two sigma-finite measure spaces states that an integrable measurable function on the product has integrable sections almost everywhere, the section integrals define Lebesgue integrable functions, and both iterated integrals equal the product integral. The integrability hypothesis is absolute integrability. For nonnegative measurable functions, Tonelli's theorem gives equality of the possibly infinite product and iterated integrals without first assuming finiteness.

The displayed function is not Lebesgue integrable. Its value at the origin can be assigned arbitrarily, since a point is a null set. On $0<x<1$ and $0<y<x/2$,Consequently Tonelli's theorem gives the divergent lower boundThe signed iterated integrals illustrate why the hypothesis matters. For $x>0$, $f(x,y)={\partial }_y[y/(x+y{)}^2]$, so integration first in $y$ and then in $x$ gives $1/2$. Integrating first in $x$ gives $-1/2$, using $f(x,y)=-{\partial }_x[x/(x+y{)}^2]$. Exceptional boundary sections have measure zero. Thus finite iterated values do not establish product integrability.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

The partial sums $s_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase to an extended-valued function $s$. By the monotone convergence theorem,A nonnegative function with finite Lebesgue integral cannot equal $+\infty$ on a set of positive measure: otherwise its integral would exceed $M$ times that measure for every $M>0$. Thus the series converges to a finite value almost everywhere. Redefine it to be zero on the exceptional null set if a finite-valued representative is desired. The resulting function is Lebesgue integrable, with

A property holds almost everywhere when its exceptional set, where the property fails, is a null set. The exceptional set need not be empty or countable.

A null set $E\subseteq \mathbb{R}$ has Lebesgue outer measure zero. Equivalently, for every $ϵ>0$ it can be covered by countably many open intervals whose total length is less than $ϵ$. This definition also applies before assuming $E$ measurable; every such set is Lebesgue measurable with measure zero.

For each $n\ge 1$, choose a countable interval cover of $E_n$ of total length below $ϵ/2^n$. The union of these covers is countable and covers ${\bigcup }_n{E}_n$, with total length less than ${\sum }_nϵ/2^n=ϵ$. As $ϵ$ is arbitrary, a countable union of null sets is null.

Given $ϵ>0$, cover $E_1$ by open intervals of total length less than $ϵ/2$, and cover $E_2$ similarly. Combining the two countable covers gives a cover of their union of total length less than $ϵ$. Thus $E_1\cup E_2$ is a null set.

For each $n$, let $N_n$ be the null set where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two sequences agree term by term, so their pointwise limits are equal: $f(x)=g(x)$. Therefore $f\sim g$. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.

The relation is reflexive because the exceptional set for $f=f$ is empty, and symmetric because the exceptional sets for $f=g$ and $g=f$ agree. If $f=g$ off a null set $N_1$ and $g=h$ off a null set $N_2$, then $f=h$ off $N_1\cup N_2$, which is null by part (b). Hence it is transitive as well, so equality almost everywhere is an equivalence relation.

A property holds almost everywhere when its exceptional set, where the property fails, is a null set. The exceptional set need not be empty or countable.

A null set $E\subseteq \mathbb{R}$ has Lebesgue outer measure zero. Equivalently, for every $ϵ>0$ it can be covered by countably many open intervals whose total length is less than $ϵ$. This definition also applies before assuming $E$ measurable; every such set is Lebesgue measurable with measure zero.

For each $n\ge 1$, choose a countable interval cover of $E_n$ of total length below $ϵ/2^n$. The union of these covers is countable and covers ${\bigcup }_n{E}_n$, with total length less than ${\sum }_nϵ/2^n=ϵ$. As $ϵ$ is arbitrary, a countable union of null sets is null.

Given $ϵ>0$, cover $E_1$ by open intervals of total length less than $ϵ/2$, and cover $E_2$ similarly. Combining the two countable covers gives a cover of their union of total length less than $ϵ$. Thus $E_1\cup E_2$ is a null set.