Here is the proof of the theorem. I couldn't understand how to apply Ergodic theorem in this proof. Let $X=(X_t)_{t\geq0}$ be a fractional Brownian motion with self-similar parameter $H\in(0,1)$. We know that when $H = \frac{1}{2}$, fractional Brownian motion is in fact a standard Brownian motion and hence a semimartingale. \paragraph{}Now fix the parameter $H$ and consider for $p > 0$ fixed \begin{equation} Y_{n,p}:= \sum_{j=1}^{2^n}|X_{j2^{-n}}-X_{(j-1)2^{-n}}|^p\big(2^n\big)^{pH-1} \end{equation} From self-similarity property we obtain that (3.6) has (for each n) the same law as \begin{eqnarray} \sum_{j=1}^{2^n}|2^{-nH}X_j-2^{-nH}X_{j-1}|^p\big(2^n\big)^{pH-1}&=&\sum_{j=1}^{2^n}|X_j-X_{j-1}|^p2^{-npH}\big(2^n\big)^{pH-1} \nonumber \\ &=&2^{-n}\sum_{j=1}^{2^n}|X_j-X_{j-1}|^p. \end{eqnarray} Noticing that the sequence $\big(X_k - X_{k-1}\big)_{k \in \mathbb{Z}}$ is stationary and ergodic, the ergodic theorem tells us that $$ \tilde{Y}_{n,p}:=2^{-n}\sum_{j=1}^{2^n}|X_j-X_{j-1}|^p\rightarrow E|X_1-X_0|^p=:\gamma_p \quad (n\rightarrow\infty) $$ almost surely and in $L^1$. Hence,
$$Y_{n,p}\overset {d}{\rightarrow}\gamma_p \quad (n\rightarrow\infty)$$
and therefore $Y_{n,p}\overset {P}{\rightarrow}\gamma_p$.Hence, \begin{equation} V_{n,p}:=\sum_{j=1}^{2^n}|X_{j2^{-n}}-X_{j-1(2^{-n})}|^p\rightarrow \left\{ \begin{array}{l l} 0 & \quad \text{if} \quad pH > 1 \\ \infty & \quad \text{if} \quad pH < 1 \end{array} \right. \end{equation} If $H > \frac{1}{2}$, we can choose $p\in(H^{-1},2$) such that $V_{n,p}\rightarrow 0$ in probability, and therefore almost surely down a fast subsequence. This implies that the quadratic variation of $X$ is zero, and so (if $X$ were to be a semimartingale) $X$ must be a finite-variation process. But since for $p\in(1,H^{-1})$,$V_p := \lim_{n\rightarrow\infty} V_{n,p}$ is almost surely infinite, and (by scaling) the p-variation on any interval is infinite almost surely, $X$ can not be finite variation. If $H <\frac{1}{2}$, we can choose $p > 2$ such that $pH < 1$, and the p-variation of $X$ on $[0,1]$ (and hence on any fixed interval) must be infinite. This contradicts the almost-sure finiteness of the quadratic variation of $X$, assuming $X$ is a semimartingale. In either way, if $H\neq\frac{1}{2}$, $X$ is not a semimartingale.$\square$