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Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact (see this notes by Kallianpur). Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function f we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact (see this notes by Kallianpur). Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $${\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)}\,.$$

Let $(\mu^n$) be a sequence of regular measures on $X$. If for any $\varepsilon > 0$ there is a metrizable compact subset $K_\varepsilon \subset X$ such that $$\inf_{n} \mu^n(K_\varepsilon) > 1 - \varepsilon\,,$$ then the sequence $(\mu^n)_n$ admits a weakly convergent subsequence $(\mu^{n_k})_k$ (see Theorem 8.6.7 in V.I. Bogachev's "Measure Theory V.2."). Hence, there is a prob. measure $\mu$ such that for any weak continuous map $f$ $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function $f$ we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact. Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function f we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact (see this notes by Kallianpur). Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function f we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

Consider a Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact. Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function f we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.

Consider a separable Banach space $X$. The Borel $\sigma$-algebra $\mathcal{B}$ of $X$ is the same when taken with respect to the weak or strong topology. Hence, the space of probability measures over $X$ is the same with respect to the weak or the strong topology.

What differs is the topology of this space right? The weak convergence of a sequence of measures $\mu^n$ to $\mu$ is given by the convergence of integrals $$\int_X f(x) \mu^n(dx) \to \int_X f(x) \mu(dx)$$ for all weak-continuous functions or for all strong-continuous functions.

Now, if $(\mu^n)$ is a tight family of probability measures on $X$ with respect to the weak topology, then $\mu^n$ is relatively compact. Hence, there is a weakly convergent subsequence to a probability measure $\mu$, i.e. for all weak continuous f $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx)\,.$$

My question is: In what cases can I show that for a sequentially continuous function f we have $$\int_X f(x) \mu^{n_k}(dx) \xrightarrow{k\to\infty} \int_X f(x) \mu(dx) \,?$$

I was trying the following: For fixed $\varepsilon > 0$, find $B_R = \{x \in X | \|x\| \leq R\}$ such that $\mu^n(B_R) > 1 - \varepsilon$ for all $n > 0$.

I was trying to restrict $\mu^{n_k}$ to $B_R$ and show convergence there, but I run into problems when dealing with $\mu^{n_k}(B_R) \to \mu(B_R)$. With respect to the strong topology I can pick R slightly larger avoiding any atom in the boundary of $B_R$, i.e. there are countably many $R$ dense in $[0, \infty)$ such that $\mu(\partial B_R) = 0$, hence $B_R$ is a continuity set. This is not possible with respect to the weak topology because the boundary of $B_R$ is $B_R$ itself.

My intuition is that a sequence of measure can not capture the difference between sequential continuous and continuous functions, but I can put that down into rigorous math.