1

\(G\) is a linear functional due to linearity of integration. To see that \(G\) is bounded we can check that \(G(f) = \int fg\ d\mu = \left\lVert fg\right\rVert_1 \leq \left\lVert f\right\rVert_p\left\lVert g\right\rVert_q\) by Hölder’s inequality, which also shows the bound \(\left\lVert G\right\rVert\leq\left\lVert g\right\rVert_q\).

To finish the proof we choose \[ f = c\cdot\operatorname{sgn}(g)\cdot\left\lvert g\right\rvert^{q-1} \] where \(c = {\left\lVert g\right\rVert_q}^{-q/p}\), note that \(\frac{q}{p} = q-1\).

First we show \(f\) is \(L^p\) and \(\left\lVert f\right\rVert_p = 1\), \[\begin{align*} {\left\lVert f\right\rVert_p}^p &= \int \left\lvert f\right\rvert^p\ d\mu \\ &= c^p \int \left\lvert g\right\rvert^{p(q-1)}\ d\mu \\ &= c^p \int \left\lvert g\right\rvert^{q}\ d\mu \\ {\left\lVert f\right\rVert_p} &= c\cdot{\left\lVert g\right\rVert_q}^{q/p} \\ &= 1 \end{align*}\] then computing \(G(f)\) we have \[\begin{align*} G(f) &= \int fg\ d\mu \\ &= c \int \left\lvert g\right\rvert^p\ d\mu \\ &= {\left\lVert g\right\rVert_q}^{1-q}\cdot{\left\lVert g\right\rVert_q}^q \\ &= \left\lVert g\right\rVert_q \end{align*}\] which shows the equality in \(\left\lVert G\right\rVert = \left\lVert g\right\rVert_p\).

2

(a)

\(\nu(\emptyset) = 0\) as \(G\) is bounded.

Let \(E_1,E_2,\dots \in \mathcal{X}\) be disjoint, as \(X\) is a finite measure space only finitely many sets have nonzero measure, so additivity of \(\nu\) carries over from linearity of \(G\).

(b)

Use Hahn decomposition theorem to express \(X = P \sqcup N\) where \(P\) is positive for \(\nu\) and \(N\) negative. Define \(\nu^+(E) = \nu(E\cap P)\) and \(\nu^-(E) = -\nu(E\cap N)\). Positivity of each measure comes from the definitions, while definiteness and countable additivity inherits from \(\nu\).

(c)

Suppose \(\mu(E) = 0\), then \(\nu^{+}(E) = \nu(E\cap P) = G(\chi_{E\cap P}) = 0\) because \(\left\lVert\chi_{E\cap P}\right\rVert = 0\), so \(\nu^+ \ll \mu\), similarly we can show \(\nu^- \ll \mu\).

Applying Radon-Nikodýn theorem, let \(g^+, g^-: X\to [0,\infty)\) such that for any \(A\in\mathcal{X}\), \[\begin{align*} \nu^+(A) &= \int_A g^+\ d\mu, \\ \nu^-(A) &= \int_A g^-\ d\mu. \end{align*}\]

(d)

For \(g^+\) we restrict our attention to measurable subsets of \(P\), let \(E\in \mathcal{X}, E\subseteq P\), then \[\nu^+(E) = \nu(E) = G(\chi_E) = \int \chi_E g^+\ d\mu \] so by linearity for all nonnegative simple functions \(\varphi\) with support in \(P\), we have \[ G(\varphi) = \int \varphi g^+\ d\mu \] then by Monotone Convergence we can generalise this equality to apply to all (nonnegative) \(f\in L^p\) with support in \(P\) \[ G(f) = \int f g^+\ d\mu \] this can be generalised to all \(f\in L^p\) with support in \(P\) by applying linearity of \(G\), then we can use a similar trick to Q1 to produce a \(f\) satisfying \(G(f)\geq \left\lVert g^+\right\rVert_q\).

Case of \(g^-\) is similar, except that \(G(f) = -\int fg^-\ d\mu\).

(e)

Let \(g = g^+ - g^-\), then \[\begin{align*} \int fg\ d\mu &= \int fg^+\ d\mu - \int fg^-\ d\mu \\ &= \int (f\cdot\chi_P)g^+\ d\mu - \int (f\cdot\chi_N)g^-\ d\mu \\ &= G(f\cdot\chi_P) + G(f\cdot\chi_N) \\ &= G(f\cdot\chi_P + f\cdot\chi_N) = G(f) \end{align*}\] seeing that \(\left\lVert g\right\rVert_q = \left\lVert G\right\rVert\) is basically Q1 again.

(f)

Suppose \(g'\) also represents \(G\), without loss of generality let \[ E = \left\{x: g(x) > g'(x)\right\} \] be non-null, let \(E_\epsilon\) denote \(\left\{x: g(x) > g'(x) + \epsilon\right\}\) and as \(E = \bigcup_{n=1}^\infty E_{1/n}\) we have \(E_\epsilon\) non-null for some \(\epsilon\). As our measure space is finite we can see that \(\chi_{E_\epsilon}\) is in \(L^p\) and \[ \int \chi_{E_\epsilon} g\ d\mu \ne \int \chi_{E_\epsilon} g'\ d\mu \] which is a contradiction.

3

(a)

As \(f = \operatorname{sgn}(f)\left\lvert f\right\rvert\) we can just show the case for nonnegative \(\left\lvert f\right\rvert\).

(b)

Let \(\varphi_n\) be a sequence of simple functions that converge to \(f\) almost everywhere.

(c)

Applying Egorov theorem as \(\lambda([a,b])\) is finite, there exists \(E_\delta\subseteq [a,b]\) with \(\lambda(E_\delta) < \delta/3\) where \(\varphi_n\) converges uniformly to \(f\) on \([a,b]\setminus E_\delta\).

(d)

Fix \(\delta > 0\), For each \(\varphi_n\), express it as \[ \varphi_n(x) = \sum_{j=1}^N a_{n,j} \chi_{E_{n,j}} \] where \(E_{n,j}\) are disjoint.

As each \(E_{n,j}\) is measurable by approximating it with a closed set then approximating the resulting complement with an open set, we can find \(G_{n,j}\) where \(\lambda(G_{n,j}) < \frac{2^{-n}\delta/3}{N}\) and \(\varphi_n\) is constant on \(E_{n,j}\setminus G_{n,j}\).

As each \(E_{n,j}\setminus G_{n,j}\) is compact their pairwise Hausdorff distance is nonzero, then \(\varphi_n\) is continuous on \([a,b]\setminus G_n\) where \(G_n = \bigcup_{n=1}^N G_{n,j}\) and \(\lambda(G_n) < 2^{-n}\delta/3\). Then \(\tilde{G}_\delta = \bigcup_{n=1}^\infty G_n\) has measure less than \(\delta/3\).

(e)

As both \(E_\delta\) and \(\tilde{G}_\delta\) are Lebesgue measurable we approximate their union with an open set to get \(G_\delta\) which contains both and has measure less than \(\delta\).

(f)

For each \(n\) consider \(\tilde{\varphi}_n = \varphi_n \cdot \chi_{G_\delta}\).

(g)

We have now shown that on \([a,b]\setminus G_\delta\), the convergence \(\varphi_n \to f\) is uniform, so \(f\) on \(\chi_{G_\delta}\) is a uniform limit of continuous functions and therefore continuous.