LHS is convergent iff RHS is convergent. Why? Achim Klenke Probability theorem 6.7

Question

The theorem goes: Let $A_{1}, A_{2} ... \in \mathcal{A}$ with $A_{N}$ increasing to $\Omega$ and $\mu (A_{N}) < \infty$ for all $N \in \mathbb{N}$. For measurable $f, g: \Omega \xrightarrow{} E$ where $E$ is a metric space, define

$\tilde{d}(f, g) := \sum_{N = 1}^{\infty} \frac{2^{-N}}{1 + \mu(A_{N})} \int_{A_{N}} \text{min}\{1, d(f(\omega), g(\omega))\} d\mu$.

Then $\tilde{d}$ is a metric that induces convergence in measure: if $f, f_{1}, ...$ are measurable, then $f_{n} \xrightarrow{} f$ in measure iff $\tilde{d}(f, f_{n}) \xrightarrow{} 0$.

In the proof, the author defines $\tilde{d}_{N}(f, g) = \int_{A_{N}} \text{min}\{1, d(f(\omega), g(\omega))\} d\mu$.

He says $\tilde{d}(f, f_{n}) \xrightarrow{} 0 $ iff $\tilde{d}_{N}(f, f_{n}) \xrightarrow{} 0 $ for all $N$. Why do we have this? First taking the infinite sum then taking the limit is the same as first taking the limit then sum? How to justify this?

Answer 1

Hint: $$\tilde{d}(f, g) $$ $$ = \sum_{N = m+1}^{\infty} \frac{2^{-N}}{1 + \mu(A_{N})} \int_{A_{N}} \text{min}\{1, d(f(\omega), g(\omega))\} d\mu $$ $$+\sum_{N = 1}^{m} \frac{2^{-N}}{1 + \mu(A_{N})} \int_{A_{N}} \text{min}\{1, d(f(\omega), g(\omega))\} d\mu$$ and the first term is less than $\sum_{N = m+1}^{\infty} \frac{2^{-N}}{1 + \mu(A_{N})}<2^{-m}$ since $1+\mu (A_N) \geq 1$. I hope you can complete the proof using this.

Answer 2

Thank @geetha290krm for your hint!

Following this hint, fix $m \in \mathbb{N}$, we have that

$\tilde{d}(f, f_{n}) \leq \frac{1}{2^{m}} + \sum_{N = 1}^{m}\frac{2^{-N}}{1 + \mu(A_{N})} \tilde{d}_{N}(f, f_{n})$.

Then, if all $\tilde{d}_{N}(f, f_{n})$ converges to zero, we have the limit of RHS is $1/2^{m}$. So by the definition of limit, for such $m$, there is a $k \in \mathbb{N}$ s.t. if $n \geq k$, we have

RHS $\leq 1/2^{m} + 1/2^{m}$ where the first $1/2^{m}$ is the limit and the second serves as the $\epsilon$.

Then for $n \geq k$, LHS also is smaller than $1/2^{m} + 1/2^{m}$. $m$ can be arbitrarily large so the limit of LHS is zero.

If not all $\tilde{d}_{N}$ converges to zero then it's easy to see $\tilde{d}$ doesn't go to zero.