Riemann surface
matlab

2.1 Example Sheet 2: questions

Problem 1. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be absolutely continuous, in the sense that
$$f(t)=f(0)+\int_0^t f^{\prime}(s) d s \text { for all } t \geq 0$$
for an integrable function $f^{\prime}$. Let $v_f(0, t)$ be the total variation of $f$ on $(0, t]$. Show that
$$v_f(0, t)=\int_0^t\left|f^{\prime}(s)\right| d s$$

Solution. Let $f_{ \pm}^{\prime}$ be the positive and negative parts of $f^{\prime}$, i.e.
$$f_{+}^{\prime}=\max \left(f^{\prime}, 0\right), \quad f_{-}^{\prime}=-\min \left(f^{\prime}, 0\right) .$$
Let $\mu_{ \pm}$be the Borel measures on $(0, \infty)$ with densities $f_{ \pm}^{\prime}$ :
$$\mu_{ \pm}(A)=\int_A f_{ \pm}^{\prime}(x) d x$$
and let $\mu=\mu_{+}-\mu_{-}$. The positive measures $\mu_{ \pm}$are finite, and supported on the disjoint sets $\left{f^{\prime}>0\right}$ and $\left{f^{\prime}<0\right}$ respectively, so $\mu$ is a signed measure, and has (unique) Hahn-Jordan decomposition $\mu_{+}-\mu_{-}$. The total variation is then $|\mu|=\mu_{+}+\mu_{-}$, which is given by \begin{aligned} |\mu(A)| & =\int_{A \cap\left{f^{\prime}>0\right}} f^{\prime}(s) d s+\int_{A \cap\left{f^{\prime}<0\right}}\left(-f^{\prime}\right)(s) d s \ & =\int_A\left|f^{\prime}\right|(s) d s \end{aligned} and \begin{aligned} \mu([0, t]) & =\int_{[0, t] \cap\left{f^{\prime}>0\right}} f^{\prime}(s) d s-\int_{[0, t] \cap\left{f^{\prime}<0\right}}\left(-f^{\prime}\right)(s) d s \ & =\int_0^t f^{\prime}(s) d s=f(t)-f(0)=: g(t) . \end{aligned}
The total variation $v_g$ of $g(t)=f(t)-f(0)$ is the same as $v_f$, and by a proposition in lectures,
$$v_g(0, t)=|\mu|(0, t)=\int_0^t\left|f^{\prime}\right|(s) d s$$

Problem 2. Let $f, g:[0, \infty) \rightarrow \mathbb{R}$ be bounded and measurable, and let $a:[0, \infty) \rightarrow \mathbb{R}$ be continuous and of finite variation. Show that
$$f \cdot(g \cdot a)=(f g) \cdot a$$
where · denotes the Lebesgue-Stieltjes integral.

Solution. Let us fix $g$, and consider
$$V_g={f: f \text { is bounded and measurable, and (6) holds }}$$
Consider $f=1[0, s]$, for some $s \in[0, \infty)$. Then
$$(f \cdot(g \cdot a))(t)=(g \cdot a)(t \wedge s)$$
and
\begin{aligned} ((f g) \cdot a)(t) & =\int_{[0, t]} g(u) 1(u \leq s) \mu_a(d u)=\int_{[0, t] \cap[0, s]} g(u) \mu_a(d u) \ & =(g \cdot a)(t \wedge s) . \end{aligned}
It follows that $1_A \in \mathcal{V} V_g$, for all $A \in \mathcal{A}={[0, t]: t \geq 0}$. The space $V_g$ is clearly linear, by linearity of the integrals on both sides of (6).

I claim that $V_g$ is a monotone class. Suppose $0 \leq f_n \uparrow f$ pointwise, $f_n \in \mathcal{V}g$ and fix $t \geq 0$. Let us write $\mu$ for the signed measure corresponding to $g \cdot a$, and $v$ for the signed measure corresponding to $a$, and $g^{ \pm}$ for the positive and negative parts. We write out $$\int{[0, t]} f_n(s) \mu_{+}(d s)-\int_{[0, t]} f_n(s) \mu_{-}(d s)=\int_{[0, t]} f_n(s) g(s) v_{+}(d s)-\int_{[0, t]} f_n(s) g(s) v_{-}(d s)$$
and consider each term separately. For the two terms on the left-hand side, we use the monotone convergence theorem for positive measures to see that
$$\int_{[0, t]} f_n(s) \mu_{ \pm}(d s) \rightarrow \int_{[0, t]} f(s) \mu_{ \pm}(d s)$$
both of which are finite, since $f$ is bounded and $(g \cdot a)$ is of total variation. On the right-hand side, $v_{ \pm}$ are finite measures on $[0, t]$, and $f_n g \rightarrow f g$ everywhere, with $f_n g$ uniformly bounded. Using dominated convergence, we see that
$$\int_{[0, t]} f_n(s) g(s) v_{ \pm}(d s) \rightarrow \int_{[0, t]} f(s) g(s) v_{ \pm}(d s)$$
and the limit on the right-hand side is finite. Reassembling, we have shown that
$$\int_{[0, t]} f(s) \mu(d s)=\int_{[0, t]} f(s) g(s) v(d s)$$
for all $t$, which is precisely the statement that (6) holds for $f$.
It therefore follows, by the monotone class theorem, that ${ } V_g$ contains all bounded measurable functions, for any bounded, measurable $g$, and we are done.

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