Past exam of the options II course of the University of Oxford / 2013 / D4

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.

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.

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 $n\ge 1$, 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.

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.

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.

Exactly those integrable functions that vanish almost everywhere:Thus this expression is a seminorm on actual functions, becoming a norm after the quotient in part (ii).

Let $N$ be the vector subspace of integrable functions that vanish almost everywhere. The Lebesgue space is the quotient vector space $L^1(\mathbb{R})={\mathcal{L}}^1(\mathbb{R})/N$: its elements are equivalence classes of functions equal almost everywhere. Addition and scalar multiplication are $[f]+[g]=[f+g]$ and $c[f]=[cf]$.

Set $\Vert [f]{\Vert }_1=\int |f|$. This is well-defined because replacing $f$ by an equal-almost-everywhere representative does not change the integral. It is nonnegative, homogeneous, and obeys the triangle inequality by $|f+g|\le |f|+|g|$. Part (i) says its only zero element is the zero class. Hence it is a norm on $L^1$.

For each integer $m\ge 1$, list the $m^2$ functionsin increasing $j$, and concatenate the blocks in increasing $m$. Each function has $L^1$ norm $1/m$, so the resulting sequence has norm tending to zero. Every $x\in (0,1)$ lies in exactly one interval of each block, where the function takes value $m$. ThereforeThis unbounded typewriter sequence with vanishing L1 norm is a moving-spike construction: the support shrinks fast enough to beat the increasing height.

No such sequence exists. By Fatou's lemma on $(0,1)$,But the proposed nonnegative measurable integrand is positive at every point of $(0,1)$, so its integral cannot be zero. Explicitly, the sets where it is at least $1/k$ cover $(0,1)$; at least one has positive measure, forcing a positive integral. This contradicts the displayed bound.