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.



Uniform boundedness principle for almost surely converging sequence of operators
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. $x$.


