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I'd like to do the following: I consider a separable Banach space $X$ with a probability measure $\mu$ on the Borel $\sigma$-algebra $\mathcal B(X)$. Additionally, I have a sequence of measurable, linear and bounded operators $T_n: X\to X$ which converges $\mu$-almost-surely, i.e. there is a set $A\in \mathcal B(X)$ with $\mu(A) = 1$ such that $T_n(x) \to T(x)$ for all $x\in A$. I also know that $T$ is a linear and continuous operator (in particular, in this case $T(x) = x$, but I don't know whether that is relevant).

The question is whether I can now apply the uniform boundedness principle/Banach-Steinhaus in order to prove that the sequence of operator norms $(||T_n||)_n$ is bounded. This would be true, if the convergence held pointwise everywhere but I wonder whether the restriction on a set of measure $1$ destroys this proof.

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    $\begingroup$ Well, you can just take a point measure. $\endgroup$ Commented May 14, 2021 at 9:45
  • $\begingroup$ All right, I see how it fails in this case. What if the measure is a non-degenerate Gaussian probability measure, or more general, a measure with full support? $\endgroup$ Commented May 14, 2021 at 10:05
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    $\begingroup$ The sequence is uniformly bounded if the set $\{x\in X: \|T_n(x)\|\text{ bounded}\}$ is of second category (=not meager) -- but this condition is quite different than having full measure. $\endgroup$ Commented May 14, 2021 at 11:06
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    $\begingroup$ @JochenWengenroth: Yes indeed. In fact, every Borel probability measure $\mu$ on a separable Banach space is Radon and so there is a $\sigma$-compact set $A$ with $\mu(A)=1$, but in an infinite-dimensional Banach space every $\sigma$-compact set is meager. So the almost everywhere convergence cannot by itself produce convergence on a non-meager set. $\endgroup$ Commented May 14, 2021 at 15:05
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    $\begingroup$ If you are interested in this general direction, I suggest that you look at Fonf, V. P.; Johnson, W. B.; Pisier, G.; Preiss, D. Stochastic approximation properties in Banach spaces. Dedicated to Professor Aleksander Pełczyński on the occasion of his 70th birthday. Studia Math. 159 (2003), no. 1, 103–119. $\endgroup$ Commented May 14, 2021 at 22:00

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No, it's not true.

Let's take $X$ to be a Hilbert space for simplicity. I will produce a non-degenerate Gaussian measure $\mu$ on $X$, and a sequence of bounded linear functionals $f_n \in X^*$ such that $f_n(x) \to 0$ for $\mu$-a.e. $x$, but $\|f_n\|_{X^*} \to \infty$. You can turn this into the desired counterexample by taking the rank-one operators $T_n x = f_n(x) x_0$ for any fixed nonzero $x_0 \in X$, or take $T_n x = x + f_n(x) x_0$ if you want $T_n \to I$ almost surely.

Let $e_i$ be an orthonormal basis for $X$, and let $\xi_i$ be an iid sequence of standard normal random variables on some probability space $(\Omega, P)$. Then one can show that the $X$-valued random series $\sum_{i=1}^\infty 2^{-i} \xi_i e_i$ converges a.s. and in $L^2(\Omega, X)$ to an $X$-valued random variable $\xi$. Let $\mu$ be the law of $\xi$; it is a Gaussian measure which is non-degenerate because $\{e_i\}$ has dense span.

Now let $f_n(x) = \frac{1}{n}\sum_{i=1}^n 2^i \langle e_i, x \rangle_X$. Since the $e_i$ are orthonormal, you may easily compute that $\|f_n\|^2_{X^*} = \frac{1}{n^2} \sum_{i=1}^n 2^{2i} \ge \frac{2^{2n}}{n^2} \to \infty$. But $f_n(\xi) = \frac{1}{n} \sum_{i=1}^n \xi_i \to 0$ a.s. by the strong law of large numbers, which is to say that $f_n(x) \to 0$ for $\mu$-a.e. $x$.

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