In Protter's book, the definition of semimartingale is conducted by a functional approach:
Definition:(total semimartingale) A process $X$ is a total semimartingale if $X$ is cadlag, adapted and $I_x:S_u \rightarrow L^0$ is continuous, where $S_u$ is the space of the simple predictable process topologized with uniform convergence and $L^0$ is the finite valued random variable topologized with the convergence in probability.
Definition:(semimartingale) A process $X$ is called a semimartingale if, for each $t\in[0,\infty)$, $X^t$ is a total semimartingale.
Following this definition, I want to understand the following corollary 2 on p.55.
Corollary 2: A local martingale with continuous paths is a semimartingale.
The proof in the book claim that this corollary 2 is a immediate consequence of the Corollary 1, which is stated as follows
Corollary 1: Each cadlag, locally square integrable local martingale is a semimartingale.
However, I don't see the connection between a local martingale with continuous paths and locally square integrability.
Further, why does both corollary only induce the process $X$ to be a semi-martingale and not a total semi-martingale? What could go wrong with the total semi-martingale?
Re Question 1: Let $(X_t)_{t \geq 0}$ be a stochastic process with continuous sample paths. If we define stopping times by
$$\tau_k := \inf\{t \geq 0; |X_t| \geq k\}, \qquad k \in \mathbb{N}$$
then the continuity of the sample paths of $(X_t)_{t \geq 0}$ implies that the following properties hold:
It follows from the very definition that $(X_t)_{t \geq 0}$ is locally square integrable.
Re Question 2: Let $(X_t)_{t \geq 0}$ be a Brownian motion and consider
$$H_n(t) := \frac{1}{\sqrt{n}} 1_{[0,n]}(t).$$
Then $H_n \to H := 0$ uniformly but
$$I_X(H_n) = \frac{X_n}{\sqrt{n}}$$
does not converge to $I_X(H)=0$ as $n \to \infty$. This shows that $I_X: S_u \to L^0$ is not continuous.