Proof. It is easy to check that \(d\) is a metric.

Let us now verify that \(X_n \overset{\mathbb{P}}{\longrightarrow}X\) implies \(d(X_n, X) \to 0.\) Suppose that \(X_n \overset{\mathbb{P}}{\longrightarrow}X\) and choose an arbitrary \(\varepsilon\in (0,1].\) Then \[\begin{gathered} d(X_n, X) = \mathbb{E}[\lvert X_n - X \rvert \wedge 1] = \mathbb{E}[\lvert X_n - X \rvert \, \mathbf 1_{\lvert X_n - X \rvert \leqslant\varepsilon}] + \mathbb{E}[(\lvert X_n - X \rvert \wedge 1) \, \mathbf 1_{\lvert X_n - X \rvert > \varepsilon}] \\ \leqslant\varepsilon+ \mathbb{P}(\lvert X_n - X \rvert > \varepsilon) \longrightarrow \varepsilon\,, \end{gathered}\] by assumption. Since \(\varepsilon> 0\) was arbitrary, we conclude that \(d(X_n, X) \to 0.\)

Conversely, suppose that \(d(X_n, X) \to 0.\) Then for all \(\varepsilon\in (0,1]\) we have, by Chebyshev’s inequality, \[\mathbb{P}(\lvert X_n - X \rvert > \varepsilon) \leqslant\frac{1}{\varepsilon} \mathbb{E}[\lvert X_n - X \rvert \wedge 1] \rightarrow 0\,,\] i.e. \(X_n \overset{\mathbb{P}}{\longrightarrow}X.\)

All that remains, therefore, is to show that the metric space \((L^0,d)\) is complete. To that end, let \((X_n)\) be a Cauchy sequence for \(d(\cdot, \cdot).\) Choose a subsequence \(Y_k = X_{n_k}\) such that \(d(Y_k, Y_{k+1}) \leqslant 2^{-k}.\) We then use the Borel-Cantelli lemma (see also Remark 3.18) with \[\mathbb{E}\Biggl[\sum_{k = 1}^\infty (\lvert Y_{k+1} - Y_k \rvert \wedge 1)\Biggr] \leqslant\sum_{k = 1}^\infty 2^{-k} < \infty\,,\] so that \[\sum_{k = 1}^\infty (\lvert Y_{k+1} - Y_k \rvert \wedge 1) < \infty \quad \text{a.s.}\,,\] which implies \[\sum_{k = 1}^\infty \lvert Y_{k+1} - Y_k \rvert < \infty \quad \text{a.s.}\,.\] Defining \[X :=Y_1 + \sum_{k = 1}^\infty (Y_{k+1} - Y_k)\,,\] we therefore have \(Y_k \overset{\text{a.s.}}{\longrightarrow} X\) as \(k \to \infty.\) Hence, \[d(Y_k, X) = \mathbb{E}[\lvert Y_k - X \rvert \wedge 1] \longrightarrow 0\] as \(k \to \infty,\) by dominated convergence. We conclude that \(d(X_n, X) \to 0\) as \(n \to \infty.\)

Proof. Part (ii) was already established in the proof of Proposition 4.3. For part (i), if \(X_n \overset{\text{a.s.}}{\longrightarrow} X\) then \(\mathbb{P}(\lvert X_n - X \rvert > \varepsilon) = \mathbb{E}[\mathbf 1_{\lvert X_n - X \rvert > \varepsilon}] \to 0\) by dominated convergence, and if \(X_n \overset{L^p}{\longrightarrow} X\) then \(\mathbb{P}(\lvert X_n - X \rvert > \varepsilon) \leqslant\frac{1}{\varepsilon^p} \mathbb{E}[\lvert X_n - X \rvert^p] \to 0\) for any \(\varepsilon> 0.\)

Proof. We proceed by contradiction and suppose that \(X_n \overset{\mathbb{P}}{\longrightarrow} X\) but \(X_n\) does not converge to \(X\) in law. The latter means that there exists \(\varphi \in C_b\) such that \(\mathbb{E}[\varphi(X_n)] \not\to \mathbb{E}[\varphi(X)].\) Hence, there exists a subsequence \((n_k)\) and \(\varepsilon> 0\) such that \[\tag{4.1} \bigl\lvert \mathbb{E}[\varphi(X_{n_k})] - \mathbb{E}[\varphi(X)] \bigr\rvert \geqslant\varepsilon\] for all \(k.\) Moreover, by Proposition 4.4 (ii), there exists a further subsequence \((n_{k_l})\) such that \(X_{n_{k_l}} \to X\) a.s. as \(l \to \infty.\) But by dominated convergence, we have \[\lvert \mathbb{E}[\varphi(X_{n_{k_l}})] - \mathbb{E}[\varphi(X)] \rvert \to 0\] as \(l \to \infty,\) in contradiction to (4.1).

Proof. The implications (i)\(\Rightarrow\)(ii) and (i)\(\Rightarrow\)(iii) are obvious. We shall show (ii)\(\Rightarrow\)(i) and (iii)\(\Rightarrow\)(ii).

To show (ii)\(\Rightarrow\)(i), suppose (ii). Let \(\varphi \in C_b.\) Choose a sequence \(f_k \in C_c\) such that \(0 \leqslant f_k \leqslant 1\) and \(f_k \uparrow 1\) as \(k \to \infty\) (you can take for instance \(f_k(x) = (1 - \lvert x/k \rvert)_+\)). Then we telescope \[\begin{aligned} \int \varphi \, \mathrm d\mu_n - \int \varphi \, \mathrm d\mu &= \int \varphi \, \mathrm d\mu_n - \int \varphi f_k \, \mathrm d\mu_n \\ &\quad+ \int \varphi f_k \, \mathrm d\mu_n - \int \varphi f_k \, \mathrm d\mu \\ &\quad+ \int \varphi f_k \, \mathrm d\mu - \int \varphi \, \mathrm d\mu\,, \end{aligned}\] and estimate each line on the right-hand side separately.

• For any \(k \in \mathbb{N}^*,\) the second line tends to zero as \(n \to \infty,\) by assumption (ii) since \(\varphi f_k \in C_c.\)

• For any \(k \in \mathbb{N}^*,\) the first line is estimated in absolute value by \[\lVert \varphi \rVert_\infty \biggl(1 - \int f_k \, \mathrm d\mu_n\biggr) \underset{n \to \infty}{\longrightarrow} \lVert \varphi \rVert_\infty \biggl(1 - \int f_k \, \mathrm d\mu\biggr)\,,\] where we again used (ii) since \(f_k \in C_c.\)

• The third line is estimated in absolute value by \[\lVert \varphi \rVert_\infty \biggl(1 - \int f_k \, \mathrm d\mu\biggr)\,.\]

Putting everything together, we conclude that for any \(k \in \mathbb{N}^*\) we have \[\limsup_{n \to \infty} \biggl\lvert \int \varphi \, \mathrm d\mu_n - \int \varphi \, \mathrm d\mu \biggr\rvert \leqslant 2 \lVert \varphi \rVert_\infty \biggl(1 - \int f_k \, \mathrm d\mu\biggr)\,.\] Since \(k\) was arbitrary, we can take \(k \to \infty,\) under which the right-hand side tends to zero by dominated convergence. This concludes the proof of (ii)\(\Rightarrow\)(i).

To show (iii)\(\Rightarrow\)(ii), suppose (iii). Let \(\varphi \in C_c.\) Choose a sequence \(\varphi_k \in H\) such that \(\lVert \varphi_k - \varphi \rVert_{\infty} \to 0\) as \(k \to \infty.\) Then for any \(k \in \mathbb{N}^*\) we have, again by telescoping, \[\begin{aligned} &\limsup_{n \to \infty} \biggl\lvert \int \varphi \, \mathrm d\mu_n - \int \varphi \, \mathrm d\mu \biggr\rvert \\ &\quad \leqslant \limsup_{n \to \infty} \Biggl(\biggl\lvert \int \varphi \, \mathrm d\mu_n - \int \varphi_k \, \mathrm d\mu_n \biggr\rvert + \biggl\lvert \int \varphi_k \, \mathrm d\mu_n - \int \varphi_k \, \mathrm d\mu \biggr\rvert + \biggl\lvert \int \varphi_k \, \mathrm d\mu - \int \varphi \, \mathrm d\mu \biggr\rvert\Biggr) \\ &\quad \leqslant 2 \lVert \varphi - \varphi_k \rVert_\infty \underset{k \to \infty}{\longrightarrow} 0\,, \end{aligned}\] where we used that for any \(k \in \mathbb{N}^*,\) the middle term on the second line tends to zero as \(n \to \infty\) by (iii). This concludes the proof of (iii)\(\Rightarrow\)(ii).