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

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?

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.

• Thanks! I write it further based on this hint. Is that what you mean?
– Tom
Nov 12, 2022 at 16:57

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.

• Yes, this is correct. Nov 12, 2022 at 23:14