First, the following lemma establishes the oscillation properties of the diffusion (martingale) through a subinterval or slab. The expected number of ‘upcrossings’ of the density function diffusion \widehat{D}(t) across a subinterval or “slab”  \mathfrak{S}_{ab}=[D_{a},D_{b}] \subset [D_{\epsilon},\infty]
is finite for all t\in\mathbf{X}\bigcup\mathbf{Y} and the number of upcrossings on a semi-infinite interval \mathbf{\mathfrak{G}}_{a \infty}=[D_{a},\infty] is always zero. This is essentially a Doob upcrossing inequality. Let \widehat{D}(t) be the density function diffusion or martingale solution of the SDE (-) for initial data \mathscr{I}_{\epsilon}=[t=t_{\epsilon},D_{\epsilon}<0,\widehat{\mathcal{B}}_{\epsilon}=0] so that \widehat{D}(t)\in[|D_{\epsilon}|,\infty). Let [|D_{\alpha}|,|D_{\beta}|]\subset[|D_{\epsilon}|,\infty)\subset\mathbf{R}^{+} be a finite interval with |D_{\alpha}|<|D_{\beta}|. Now let \mathcal{U}(\alpha,\beta:t) denote the number of ‘upcrossings’ of the interval or slab \mathcal{S}_{\alpha\beta}=[|D_{\alpha}|,|D_{\beta}|], which is the number of times that the density function diffusion |\widehat{D}(t)| has passed from below |D_{\alpha}| to above D_{\beta}| at some t\in\mathbf{Y}\bigcup\mathbf{Z}. Then if \widehat{D}(t) is a martingale

[1] The number of upcrossings through the slab \mathfrak{S} is finite for all t\in\mathbf{Y}\bigcup\mathbf{Z}

\mathbf{E}(\mathbf{\mathcal{U}}(a \rightleftarrows b; t)~\le~\frac{1}{D_{b}-D_{a}} \mathbf{E}([|\widehat{D}(t)|]]+|D_{a}|)

=  \frac{1}{D_{b}-D_{a}}\mathbf{E}|D_{\epsilon}| +\frac{k^{1/2}}{D_{b}-D_{a}} \mathbf{E}({\displaystyle \int_{t_{\epsilon}}^{t}}|(D(v))^{2} (D(v)-1))^{1/2} \otimes d\widehat{\mathcal{B}}(v)+D_{a}) < \infty  

[2] The limit |D_{\beta}| \rightarrow \infty gives a semi-infinite
interval or slab  [|D_{\alpha}|,\infty) so that the number of upcrossings from below  D_{a} to infinity at any  t\in\mathbf{Y}\bigcup\mathbf{Z}, is zero

 \mathbf{E}(\mathbf{\mathcal{U}}(a,\infty :t)]  <=lim_{|D_{b}| \rightarrow \infty}  \frac{1}{D_{b}-D_{a}}(\mathbf{E}(\widehat{D}(t)|=0

so that for the initial data, the density diffusion  \widehat{D}(t) never blows up for any finite  t\in\mathbf{Y}\bigcup\mathbf{Z}. Define the ‘hitting times’  {S}_{n}=\inf\lbrace t>{T}_{n-1}:|D(t)|\le |D_{a}|\rbrace

{T}_{n}=\inf\lbrace t>{S}_{n}:|D(t)|\ge |D_{b}|\rbrace 

with  T_{0}=0 and n\in\mathbf{N}. Construct a stochastic process

 \widehat{Z}(t) = \sum_{n}(|\widehat{D}(t \wedge {T}_{n})|-|\widehat{D}(t \wedge  {S}_{n})|)

Beginning with the first time T_{1} that |D(t)| \le |D_{a}|, the process
\widehat{Z}(t) evolves via increments of \widehat{D}(s) until the the first time {T}_{1}\ge|D_{b}|. The process repeats if again the martingale diffusion |\widehat{D}(t)| falls below D_{a}| or |\widehat{D}(t)|<=  |D_{a}| at time \mathcal{S}_{2} and so on. All terms in the summation vanish for sufficiently large n that \mathcal{S}_{n}>t so that there are at most \mathbf{\mathcal{U}}(a\rightleftarrows\|t) terms that are non-zero. Then

  \widehat{Z}(t)=\sum_{n}\big(\widehat{D}(t\wedge{T}_{n})-\widehat{D} (t\wedge{S}_{n})\big)

\ge (D_{B}-D_{A}) \mathbf{\mathcal{U}}(a\rightleftarrows |t)(\widehat{D}(t)-|D_{a}) 

so that

 \mathbf{E}(\mathbf{\widehat{Z}}(t)D) \ge (D_{\beta}-D_{\alpha})\mathbf{E} (\mathbf{\mathcal{U}}(a \rightleftarrows~B;t) )

 +\mathbf{E}(D(s)-D_{\alpha})  \ge (D_{\beta}-D_{\alpha})\mathbf{E} \mathbf{\mathcal{U}}(a\rightleftarrows b; ~t)-\mathbf{E}(D(s)- D_{\alpha})

 \ge (D_{\beta}-D_{\alpha})\mathbf{E}\mathbf{\mathcal{U}}(a\rightleftarrows b; ~t) \mathbf{E}( D(t)|-D_{\alpha}) 

Since \widehat{Z}(t) is also a martingale then

\mathbf{E}(\widehat{Z}(t))=\mathbf{E}\langle \widehat{Z}_{o}). It follows that

\ \mathbf{E}(\mathbf{\mathcal{U}}(a\rightleftarrows b|t)~<=~\frac{1}{D_{\beta}-D_{\alpha}} (\mathbf{E}(|\widehat{D}(t)|)+|D_{\alpha} ))

Taking the limit as |D_{b}| \rightarrow \infty.

 \mathbf{E}(\mathbf{\mathcal{U}}(\alpha, \infty:t))= lim_{D_{\beta}, \infty}\mathbf{E}(\mathbf{\mathcal{U}}(a\rightleftarrows b:t)]

~\le~lim_{D_{\beta}\rightarrow\infty} \frac{1}{D_{\beta}-D_{\alpha}}(\mathbf {E}\langle \big|\widehat{D}(t)|\rangle+|D_{\alpha}\big|\big)=0 

This then gives  \mathbf{E}(\mathbf{\mathcal{U}}(a\rightarrow \infty;~ t)=0 for any t\in\mathbf{Y}\bigcup\mathbf{Z} so a blowup never occurs.

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