Regarding a proof of Tanaka's formula This is in context of a proof of Tanaka's formula. I know that the general idea of the proof is to approximate |x| by $C^2$ functions. Therefore, we defined $f_\epsilon(x) = \sqrt{\epsilon + x^2}$ and got by Ito's formula
\begin{equation}
f_\epsilon (B_t) = \sqrt{\epsilon} + \int_0^t \frac{B_s}{\sqrt{\epsilon + B_s^2}} \ dB_s + \int_0^t \frac{1}{2}\  \frac{\epsilon}{(\epsilon + B_s^2)^{\frac{3}{2}}} \ ds.
\end{equation}
Now it was proposed to show that for any $T>0$
\begin{equation}
\sup_{t\leq T} \left| \int_0^t \frac{B_s}{\sqrt{\epsilon + B_s^2}} \ dB_s - \int_0^t sgn(B_s) \ dB_s \right| \to 0 \text{   in } L^2 \text{ as } \epsilon \to 0
\end{equation}
which is not difficult to see using Ito's Isometry and Doob's inequalities. Now it is said that we can infer that there exists a continuous and increasing process $L$ s.t.
\begin{equation}
|B_t| = \int_0^t sgn(B_s)\ dB_s + L_t 
\end{equation}for all $t\geq 0$ and for any $T>0$
\begin{equation}
\sup_{t\leq T} \left| \int_0^t \frac{1}{2}\  \frac{\epsilon}{(\epsilon + B_s^2)^{\frac{3}{2}}} \ ds - L_t \right| \to 0 \text{   in } L^2 \text{ as } \epsilon \to 0.
\end{equation}
This is the part I don't get. How can we conclude that there exists such a process $L$? What does the $L^2$ convergence give us here and what do we need the supremum for?
Also, $sgn(0)= 0$ otherwise the usual definition. 
 A: For fixed $\epsilon>0$ set
$$L_t^{(\epsilon)} := \frac{1}{2} \int_0^t \frac{\epsilon}{(\epsilon+B_s^2)^{3/2}} \, ds,$$
i.e.
$$f_{\epsilon}(B_t) = \sqrt{\epsilon} + \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s+  L_{t}^{\epsilon}.$$
Since $f_{\epsilon}(B_t) \to |B_t|$ in $L^2$ and the stochastic integral converges in $L^2$ to $\int_0^t \text{sgn}(B_s) \, dB_s$, it follows that
$$L_t^{(\epsilon)} = f_{\epsilon}(B_t)-\sqrt{\epsilon} - \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s \xrightarrow[L^2]{\epsilon \to 0} |B_t| - \int_0^t \text{sgn}(B_s) \, dB_s =: L_t $$
for fixed $t \geq 0$. Note that $(L_t)_{t \geq 0}$ has (a modification with) continuous sample paths since the stochastic integral has continuous sample paths a.s. (the continuity of $|B_t|$ is obvious from the continuity of $B_t$). Now
$$\sup_{t \leq T} |L_t^{(\epsilon)}-L_t| \leq \sqrt{\epsilon}+ \sup_{t \leq T} |f_{\epsilon}(B_t)-|B_t|| + \sup_{t \leq T} \left| \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s - \int_0^t \text{sgn}(B_s) \, dB_s \right|.$$
The right-hand side converges to $0$ in $L^2$ as $\epsilon \to 0$, and so does the right-hand side. Convergence in $L^2$ implies that there exists a subsequence converging almost surely, i.e. we can find $\epsilon_k \downarrow 0$ such that $$\sup_{t \leq T} |L_t^{(\epsilon_k)}-L_t| \to 0 \quad \text{a.s.}$$ Since $t \mapsto L_t^{(\epsilon)}$ is increasing, it follows that the limit is also increasing in $t$:
$$L_s = \lim_{k \to \infty} L_s^{(\epsilon_k)} \leq \lim_{k \to \infty} L_t^{(\epsilon_k)} = L_t$$
for any $s \leq t \leq T$. 
Remark: Note that we have shown, as a by-product, that
$$|B_t| = \int_0^t \text{sgn}(B_s) \, dB_s + L_t$$
is the Doob-Meyer decomposition of the submartingale $(|B_t|)_{t \geq 0}$.
