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 Jan 20 accepted Shift Invariance of Backward Martingales for tail trivial probability measures Jan 20 comment Shift Invariance of Backward Martingales for tail trivial probability measures I was just verifying that :) Jan 20 comment Shift Invariance of Backward Martingales for tail trivial probability measures @NateEldredge I agree that different versions of the conditional expectation give us different sets ${x:g(x)=g(σx)}$ but I guess that what I had in mind (and not written here) is that independently of the choice of the version of the conditional expectation, this set always have measure one ? Jan 20 comment Shift Invariance of Backward Martingales for tail trivial probability measures This construction is nice, but I did not understand why we can apply the zero-one law since the r.v. $(X_i)_{i\in\mathbb{N}}$ are not independent. Of course, the projections $Y_i\in \{0,1\}^2$ are independent and the theorem applies, but its tail $\sigma$-algebra is different from the tail $\sigma$-algebra generated by the $X_i$ variables (which seems richer). Jan 20 awarded Inquisitive Jan 20 comment Shift Invariance of Backward Martingales for tail trivial probability measures @NateEldredge I guess your last comment answer the question. But I need to digest it because it implies that any probability measure trivial on "this" tail $\sigma$-algebra is shift invariant (its restricition to the tail $\sigma$-algebra) which seems very strong claim. Anyway thanks a lot for this clarification. Jan 19 revised Shift Invariance of Backward Martingales for tail trivial probability measures edited body Jan 19 comment Shift Invariance of Backward Martingales for tail trivial probability measures Yes. I am fixing it. Thanks. Jan 19 asked Shift Invariance of Backward Martingales for tail trivial probability measures Jan 4 revised Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$? Improved formatting and clarity of the question. Dec 22 comment Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$? Thanks again @ChristianRemling but I don't understand why. I will think about it more. Dec 21 comment Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$? Dear @ChristianRemling first of all thanks for your comment. From the hypothesis it follows that $T-(\lambda-\varepsilon)$ has bounded inverse and therefore $\exists \ c>0$ such that $c\|x\|_{\infty}\leq \| [T-(\lambda-\varepsilon)]x\|_{\infty}$. From your hint should I be able to conclude that there is a positive constant $d$ such that $d\|x\|_{L^1}\leq \| [T-(\lambda-\varepsilon)]x\|_{L^1}$ ? Could you elaborate a little bit more on your comment ? Dec 18 asked Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$? Jul 27 awarded Electorate Feb 20 awarded Necromancer Dec 8 awarded Yearling Jul 2 awarded Curious May 28 awarded Popular Question May 14 comment Is $\text{Bow}(X,T)$ a Banach Space? @BenWillson in this context $C^0(X)$ is usually the space of all continuous functions taking values on $\mathbb{R}$ or $\mathbb{C}$. May 8 awarded Popular Question