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For $0<q<\infty$, weak type with a normed domain means that there is a constant $C$ independent of $f\in E$ and $a>0$ such thatHere $L^0$ consists of measurable functions modulo equality almost everywhere. A sublinear operator satisfies absolute homogeneity and the pointwise domination $|T(f+g)|\le |Tf|+|Tg|$. In particular the definition controls the output's superlevel measures, not necessarily its strong $L^q$ norm. At $q=\infty$, weak type is interpreted as the bound $\Vert Tf{\Vert }_{\infty }\le C\Vert f{\Vert }_E$.

A useful theorem is dense approximation under a weak-type maximal bound. Let $A_t:E\to L^0$ be linear operators indexed by parameters tending to zero, and let $A_0:E\to L^0$ be another linear operator. Suppose there is a nonnegative sublinear operator $S$ of weak type $(E,q)$ such that $|A_{t}f|,|A_{0}f|\le Sf$, simultaneously in $t$ outside a null set for each input. Assume the suprema and limiting suprema used below are measurable. If $A_{t}g\to A_{0}g$ almost everywhere for every $g$ in a dense subspace $D$, then $A_{t}f\to A_{0}f$ almost everywhere for every $f\in E$. For a sequence of operators all suprema are automatically countable; continuously parametrized averages below also have measurable suprema.

To prove the theorem, fix $f$ and $g\in D$. Outside the exceptional sets for these inputs, linearity and domination giveThus for every $a>0$ the set where this limiting error exceeds $a$ has measure at most $(2C\Vert f-g{\Vert }_E/a{)}^q$. Choose a sequence $g_k\in D$ converging to $f$ in norm and take the union of their null exceptional sets. Letting $k\to \infty$ makes the bound zero. Taking countably many positive rational $a$ proves convergence almost everywhere. This proves the requested sufficient condition rather than merely invoking a convergence theorem.

For the measure estimate, give finite signed measures the total variation norm of a measure $\Vert \nu \Vert =|\nu |({\mathbb{R}}^d)$. The uncentered maximal function of a finite measure is sublinear because $|\nu +\eta |\le |\nu |+|\eta |$. Its set $E_a=\{m_u\nu >a\}$ is open: it is precisely the union of open balls whose variation mass exceeds $a$ times their volume.

Fix a compact subset $K\subset E_a$ and select a finite cover of $K$ by these high-density balls. Order that finite collection by decreasing radius. Keep a ball if it is disjoint from all previously retained balls. Every discarded ball intersects a retained ball of at least its radius and is contained in that retained ball's triple dilation. The retained balls $B_j$ are disjoint and their triples cover $K$. ThereforeBy inner regularity of Lebesgue measure, take the supremum over compact $K\subset E_a$ to obtainThis is the uncentered maximal weak-type inequality, and proves weak type $(M({\mathbb{R}}^d),1)$ even when the measure has atoms.

Finally work in dimension one. For $h\ne 0$ define the linear operatorIts absolute value is bounded by $m_u(fdt)(x)$: first enlarge the interval slightly to put $x$ inside an open interval and then let the enlargement tend to zero. The endpoint masses vanish because $fdt$ is absolutely continuous. Taking $A_{0}f=f$, both operators are dominated by $Sf=m_u(fdt)+|f|$. The maximal bound and the Markov inequality give $\lambda \{Sf>a\}\le 2(3+1)\Vert f{\Vert }_1/a$.

Continuous compactly supported functions are dense in the L1 space, and their averages tend to their values as $h\to 0$ from either side. For a fixed $x$, the interval integral varies continuously for $h\ne 0$, so the required suprema can be taken over rational $h$ and are measurable. Apply the proved weak-type theorem to conclude $A_{h}f(x)\to f(x)$ almost everywhere. Hence

Put $\mu =Du$, a finite signed Radon measure, and defineFor $\phi \in C_c^1((a,b))$, Fubini's theorem for the signed measure givesThus $D{F}_l=\mu =Du$. A distribution on a connected interval with zero derivative is constant: any compactly supported test of integral zero is the derivative of a compactly supported test, so it pairs to zero with $u-F_l$. Choose the resulting constant $c$. Thenare one-sided representatives of a one-dimensional BV function. The first equals $u$ almost everywhere. Their difference is $\mu (\{t\})$, nonzero at at most countably many points, so the second also equals $u$ almost everywhere.

Finite-measure continuity applied to $|\mu |$ proves $F_l(s)\to F_l(t)$ as $s\uparrow t$, and $F_r(s)\to F_r(t)$ as $s\downarrow t$. Moreover the opposite one-sided limits are $F_l(t+)=F_r(t)$ and $F_r(t-)=F_l(t)$. Consequently both representatives are continuous exactly where $\mu (\{t\})=0$. For each positive integer $m$, there are only finitely many measure atoms of magnitude at least $1/m$, since their total magnitudes are bounded by $|\mu |((a,b))$. Their union is countable, proving the requested discontinuity bound. This does not require the jump points to be isolated; they can be dense.

The hint's one-dimensional statement is consistent with this construction: zero ${\mathcal{H}}^0$ measure means an empty set, so $S_u\subseteq J_u$. This concerns approximate discontinuities of the BV space class; arbitrary changes of representative at points could create artificial pointwise discontinuities. The interval-mass representatives above remove that ambiguity and give the required one-sided continuity directly.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

Consider the probability mass functions $P=(0.4,0.1,0.2,0.3)$ and $Q=(0.1,0.3,0.3,0.3)$ on the finite set $S=\{A,B,C,D\}$. Their difference is $(0.3,-0.2,-0.1,0)$. ThusThe event $E=\{A\}$ attains this difference: $P(E)=0.4$ and $Q(E)=0.1$. Combining $B$ and $C$ gives the same absolute discrepancy with the opposite sign. Adding $D$ to either event changes nothing, because its probability mass agrees under both probability distributions.

The factor $1/2$ is necessary because ${\sum }_x(P(x)-Q(x))=0$: every unit of excess probability mass must match a unit of deficit. For any event, including a negative term cannot increase its signed discrepancy, while omitting a positive term cannot increase it either. Therefore $\{x:P(x)>Q(x)\}$ attains the positive maximum. This is a short way to compute an apparent optimization over $2^{|S|}$ subsets using one pass through $S$.

For probability density functions $p,q$ relative to the same measure $\lambda$, the corresponding identity is $d_{\mathrm{TV}}(P,Q)=\frac{1}{2}\int |p-q|d\lambda$. Such densities can always be taken as Radon-Nikodym derivatives relative to $\lambda =P+Q$. Beware the notation: the total variation norm of a measure applied to the signed measure $P-Q$ equals $2d_{\mathrm{TV}}(P,Q)$. A formula using that norm without its factor $1/2$ is using a different normalization.