A real-valued random variable is a measurable function $Y:(\Omega ,\mathcal{F})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$: equivalently, $\{Y\le t\}\in \mathcal{F}$ for every real $t$.

For $p=1$ the asserted inequality is equality. For $p>1$, let $q=p/(p-1)$. If $\mathbb{E}{Y}^p=\infty$, the inequality is automatic. Otherwise Hölder's inequality on the probability space givesThus $\boxed{\Vert Y{\Vert }_1\le \Vert Y{\Vert }_p}$, using that the total measure is one.

Let $U$ have the uniform distribution on $(0,1)$, set $Y=1/U$, and take $X_n=Y/n$. The variable $Y$ is finite almost surely, so $X_n\to 0$ almost surely. For any $ϵ>0$ and all sufficiently large $n$,The harmonic series diverges, soThis dependent sequence fails complete convergence. All its members are scalar multiples of the same nonconstant random variable, which explains why the independence hypothesis cannot be dropped.

Suppose first that complete convergence holds. For each positive integer $j$, apply the First Borel-Cantelli lemma to $\{|X_n-X|>1/j\}$. Outside a countable union of null events, only finitely many of these exceedances occur for every $j$. This is almost sure convergence, with no independence assumption.

For the converse, an important point is that independence of $X_n$ does not immediately imply independence of $X_n-X$. Instead use the fact that an almost-sure limit of independent random variables is constant. Indeed, $Z={\mathrm{limsup}}_n{X}_n$ is measurable for the tail sigma-algebra. The Kolmogorov zero-one law makes each event $\{Z\le q\}$, $q\in \mathbb{Q}$, have probability zero or one. Since $Z=X$ almost surely and $X$ is finite, these probabilities form the distribution function of a point mass: there is a finite constant $c$ with $X=c$ almost surely.

Now $\{|X_n-c|>ϵ\}$ are independent events. Almost sure convergence says that their infinitely-often event has probability zero. The contrapositive of the Second Borel-Cantelli lemma givesReplacing $c$ by the almost surely equal $X$ leaves each probability unchanged. Thus complete convergence and almost sure convergence are equivalent for independent sequences.

By convergence in probability, choose recursively $n_k>n_{k-1}$ so thatThe corresponding events have a summable sequence of probabilities. The First Borel-Cantelli lemma, which requires no independence, therefore makes the probability of infinitely many of them zero. On the complementary event, $|X_{n_k}-X|\le 2^{-k}$ for all sufficiently large $k$, provingThe recursively chosen indices are strictly increasing and hence tend to infinity.

Independent random variables satisfy, for every finite selection of distinct indices $n_1,\dots ,n_m$ and Borel sets $C_1,\dots ,C_m$,The Second Borel-Cantelli lemma says that if events $E_n$ are independent and ${\sum }_n\mathbb{P}(E_n)=\infty$, thenHere ${\mathrm{limsup}}_n{E}_n={\bigcap }_N{\bigcup }_{n\ge N}{E}_n$ is the event that infinitely many of the $E_n$ occur.

Fix $ϵ>0$ and use the indicator random variables $Y_n=1_{\{|X_n-X|>ϵ\}}$. Almost sure convergence gives ${\mathrm{limsup}}_n{Y}_n=0$ almost surely, while $0\le Y_n\le 1$. The Reverse Fatou lemma states that for random variables bounded above by an integrable random variable,It follows thatSince $ϵ$ was arbitrary, this proves convergence in probability. The integrable upper bound is essential to the application of the Reverse Fatou lemma.

Almost sure convergence meansThus a single event of probability one supports ordinary pointwise convergence. Convergence in probability means that for every $ϵ>0$,The exceptional event is allowed to change with $n$ in the second definition.

The difference between the high-strike and low-strike put payoffs satisfiesIf its price were negative, buying the nonnegative-payoff spread would generate an arbitrage. If its price exceeded the discounted maximum payoff, selling it and buying a bond for that maximum would generate an arbitrage. Thus the sharper inequality isFor the positive-rate convention of the paper this implies the printed upper bound $K_1-K_2$. If rates were allowed to be negative, the undiscounted upper bound need not hold; the discounted bound is the correct rate-independent statement. This distinction is about the stated financial assumptions, not a change to the transcribed inequality.

A stock share plus a put has terminal value $S_T+(K-S_T{)}^{+}=\max (S_T,K)\ge K$. If $S_t+P(t,S_t)<K{e}^{-r(T-t)}$, buy that package and borrow the price of a bond paying $K$. The package covers the debt at maturity, while the strictly positive initial surplus can be invested. This would be an arbitrage. ConsequentlyTogether with payoff nonnegativity, the lower bound can be strengthened to the maximum of this expression and zero.

The terminal put payoff obeys $0\le (K-S_T{)}^{+}\le K$. If the put cost more than the zero-coupon bond paying $K$, sell the put and buy that bond. Invest the strictly positive initial difference in the bank account. At maturity the bond covers every possible put liability and the extra investment remains positive. This arbitrage is excluded, so

If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<ϵ/2$. The triangle inequality makes its oscillation less than $ϵ$, so $\phi (x_0)=0$.

Conversely, if $\phi (x_0)=0$, then for every $ϵ>0$ some $n$ has ${\phi }_n(x_0)<ϵ$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<ϵ$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.

The continuity set of a function is thereforeEach set in the intersection is open by part (i), so this is a G-delta set and in particular a Borel set, even if $f$ itself is not measurable.

Boundedness of $f$ makes $\phi$ finite and nonnegative. Suppose $\phi (x_0)<a$. By its infimum definition, some $n$ satisfies ${\phi }_n(x_0)<a$. If $|x-x_0|<1/(2n)$, thenThe supremum over the smaller interval cannot exceed that over the larger one. HenceThus $\{\phi <a\}$ is open for every $a$, proving $\phi$ is upper semicontinuous. Measurability of the original $f$ was never needed: these are purely pointwise oscillation bounds.

For a lower semicontinuous function, each upper-ray inverse image is open, hence Borel, so part (a)(i) applies directly. For an upper semicontinuous $h$, the function $-h$ is lower semicontinuous becauseis open. Negation is continuous and Borel measurable, so $h=-(-h)$ is Borel measurable too. Thus both semicontinuity classes are Borel measurable.

Every open interval is Borel, since $(a,b)={\bigcup }_{m\ge 1}(a,b-1/m]$, omitting empty pieces. Open subsets of the line are countable unions of open intervals and hence Borel. Conversely, each half-open interval is obtained from open sets by sigma-algebra operations, so this is the usual Borel sigma-algebra.

A continuous function has open, and therefore Borel, inverse images of open sets. Since open sets generate the target Borel sigma-algebra, it is a Borel measurable function. A discontinuous example is $1_{[0,\infty )}$, whose only level sets are built from the Borel half-lines; it jumps at zero.

A Borel measurable function has $f^{-1}(B)\in \mathcal{B}$ for every Borel set $B$. Thus the upper-ray condition is necessary.

For sufficiency, note that the rays generate the stipulated sigma-algebra: $(a,b]=(a,\infty )\setminus (b,\infty )$, and $(y,\infty )={\bigcup }_{m>y}(y,m]$, with integer $m$. The familyis a sigma-algebra because inverse images preserve complements and countable unions. If it contains all upper rays, it contains their generated Borel sigma-algebra. Therefore upper-ray preimages suffice to prove Borel measurability.

In any proposed decomposition, $Z_n^{\prime }=X_n-Y_n^{\prime }$ is integrable. Its increment is ${\mathcal{F}}_{n-1}$-measurable by predictability. Taking conditional expectation of the difference equation givesThe martingale increment has conditional mean zero, and a measurable integrable variable equals its own conditional expectation. Starting from $Z_0^{\prime }=0$, summation gives $Z_n^{\prime }={\sum }_{j=0}^n{G}_j=Z_n$. Hence $Y_n^{\prime }=X_n-Z_n^{\prime }=Y_n$. Thus the normalized predictable decomposition is unique, with a common probability-one event available for all integer times by countability.

The process $Y_n=X_n-Z_n$ is adapted and integrable. Since $Z_{n+1}=Z_n+G_{n+1}$ is already ${\mathcal{F}}_n$-measurable,Therefore $Y$ is a martingale, and $X=Y+Z$ is the Doob decomposition in discrete time. Its normalization is $Z_0=0$, so the initial value is included in $Y_0=X_0$.

The supermartingale property and linearity of conditional expectation give, for $n\ge 1$,Each $G_n$ is integrable and ${\mathcal{F}}_{n-1}$-measurable, while $G_0=0$. Thus $Z_n$ is integrable andMoreover, all terms $G_j$, $j\le n+1$, are ${\mathcal{F}}_n$-measurable, so $Z_{n+1}\in m{\mathcal{F}}_n$. This is predictability of the decreasing component in the Doob decomposition theorem.

A martingale is an adapted process $X_n$ with $\mathbb{E}|X_n|<\infty$ for every $n$ andAdaptedness means that $X_n$ is ${\mathcal{F}}_n$-measurable. Iterating the tower property gives the analogous identity for any two deterministic times $m\le n$.

Let $Y,Z$ be two versions. For every $A\in \mathcal{G}$, their defining identities imply ${\int }_A(Y-Z)d\mathbb{P}=0$. The set $A_m=\{Y\ge Z+1/m\}$ belongs to $\mathcal{G}$, and thereforeEvery $A_m$ is null. Their union is $\{Y>Z\}$, so it too is null. Reversing $Y,Z$ proves the same for $\{Z>Y\}$. Hence versions of conditional expectation are equal almost surely.