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For fixed $u>0$, define the left-limit sigma-algebra by ${\mathcal{F}}_{u-}=\sigma ({\bigcup }_{s<u}{\mathcal{F}}_s)$. The section at $u$ of a generating rectangle $A\times (s,t]$ of the predictable sigma-algebra is $A$ when $s<u\le t$, and empty otherwise. In the first case $A\in {\mathcal{F}}_s\subseteq {\mathcal{F}}_{u-}$. Sections of the time-zero generators are empty for $u>0$.

The collection of subsets of sample-time space whose section at $u$ belongs to ${\mathcal{F}}_{u-}$ is a sigma-algebra: sections commute with complements and countable unions. It therefore contains all of $\mathcal{P}$. Apply this to $\{(\omega ,t):H_t(\omega )\in E\}$ for every Borel set $E$. Thus predictable sections at deterministic times satisfyAt $u=0$ the same argument gives ${\mathcal{F}}_0$-measurability, using the convention ${\mathcal{F}}_{0-}={\mathcal{F}}_0$. The PDF asks for information strictly before $u$; the TeX aid drops the minus sign.

Let $Y$ be a nonconstant Bernoulli random variable, let ${\mathcal{F}}_t$ be trivial for $t<1$, and let ${\mathcal{F}}_t=\sigma (Y)$ for $t\ge 1$. This is a right-continuous filtration after completion. The processis adapted and càdlàg. If it were predictable, its value $X_1=Y$ at the deterministic time $1$ would be measurable with respect to the left-limit sigma-algebra ${\mathcal{F}}_{1-}=\sigma ({\bigcup }_{t<1}{\mathcal{F}}_t)$, which is trivial. That contradicts the choice of $Y$, so $X$ is not predictable.

For a predictable process $H$ and deterministic $t>0$, $H_t$ is measurable for the left-limit sigma-algebra ${\mathcal{F}}_{t-}$. This follows by taking sections of the generating predictable rectangles. At time zero use ${\mathcal{F}}_{0-}={\mathcal{F}}_0$.