 Solution

The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; finite variation alone does not supply them. For example, $M=0$, $A_t=1_{\{t\ge 1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.

Define $f_n(x)={\int }_0^x\varphi (ny)dy$. Then $f_n\in C^2$, $|f_n^{\prime }|\le 1$ and $f_n^{\prime \prime }(x)=n{\varphi }^{\prime }(nx)\ge 0$. The Itô formula gives the canonical continuous semimartingale decompositionThe initial value is zero. The first term is a continuous martingale, since $M$ is bounded and the integrand is bounded. The second is continuous adapted finite variation, hence predictable. This decomposition is unique because a continuous finite-variation local martingale is constant. The phrase Doob-Meyer here must mean this semimartingale decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the Doob-Meyer decomposition theorem. Indeed $f_n^{\prime }(0)=-1$, so the bounded deterministic drift $A_t=t\wedge \delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.

The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,For a fixed horizon $T$, $\mathbb{E}[M{]}_T=\mathbb{E}{M}_T^2\le K^2$. The dominated convergence theorem for the finite measure $\mathbb{E}d[M]$ on $[0,T]$ givesThe Itô isometry, extended from simple predictable integrands by their density, and the Doob L2 maximal inequality now implyHence