## Prove that $\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)|^2>r \right\}\right)=0$

Let $$(X,\mu)$$ be a measure space and $$\phi=(\phi_1,\cdots,\phi_d)\in L^{\infty}(X)$$.

Let $$r=\max\left\{\sum_{i=1}^d|z_i|^2; (z_1,\cdots,z_d)\in \mathcal{C}(\phi)\right\},$$ where $$\mathcal{C}(\phi)$$ is consisting of all $$z = (z_1,\cdots,z_d)\in \mathbb{C}^d$$ such that for every $$\varepsilon>0$$ $$\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)-z_i|<\varepsilon \right\}\right)>0 .$$

Why $$\sum_{i=1}^d |\phi_i(t)|^2\le r$$ for $$\mu$$-almost every $$t\in X$$.

## Showing a random variable converges to $\mu$ in probability

Let $$X_{1}, \ldots X_{n}$$ be a sequence of i.i.d random variables with $$\mathbb{E}[X_{1}] = \mu$$ and $$\text{Var}(X_{1}) = \sigma^{2}$$. Let

$$Y_{n} = \frac{2}{n(n + 1)}\sum_{i = 1}^{n} iX_{i}.$$

Use Chebyshev’s Inequality to show that $$Y_{N}$$ converges to $$\mu$$ in probability as $$n \to \infty$$. That is, for every $$\epsilon > 0$$, show that $$\lim_{n\to\infty} P(|Y_{N} – \mu| > \epsilon) = 0$$.

So, I wasn’t too sure about how to approach this problem. First, I computed $$\mathbb{E}[Y_{n}]$$ as follows:

$$\mathbb{E}\left[\frac{2}{n(n + 1)}\sum_{i = 1}^{n} iX_{i}\right] = \frac{2}{n(n + 1)}\sum_{i = 1}^{n} \mathbb{E}[iX_{i}]$$

$$= \frac{2}{n(n + 1)} \left(1(\mu) + 2(\mu) + 3(\mu) + \cdots n(\mu) \right)$$

$$= \frac{2}{n(n + 1)} \cdot \mu\left(\frac{n(n + 1)}{2}\right)$$

$$= \mu.$$

Then, I computed the variance. First compute the second moment:

$$\mathbb{E}\left[Y_{n}^{2}\right] = \frac{4}{n^{2}(n + 1)^{2}} \sum_{i = 1}^{n} \mathbb{E}[i^{2} X_{i}^{2}] = \frac{4}{n^{2}(n + 1)^{2}} \cdot \left(1^2 \mu^2 + 2^2 \mu^2 + \cdots n^2 \mu^2 \right)$$

$$= \mu^{2},$$

which means that $$\text{Var}(Y_{N}) = 0.$$

I don’t know if I actually computed the variance right. I also don’t know what to do next. Any help would be appreciated. In particular, I think that Chebyshev’s Inequality states that

$$P(|Y_{N} – \mu| \geq \epsilon\} \leq \frac{\sigma^{2}}{k^{2}} = 0,$$

but since probability cannot be negative, we must have it equal to $$0$$? I don’t really know.

## $\sigma$-finite measure $\mu$ so that $L^p(\mu) \subsetneq L^q(\mu)$ (proper subset)

I’m looking for a $$\sigma$$-finite measure $$\mu$$ and a measure space so that for

$$1 \le p

$$L^p(\mu) \subsetneq L^q(\mu)$$

I tried the following:

Let $$1 \le p and $$\lambda$$ the Lebesgue measure on $$(1,\infty)$$ which is $$\sigma$$-finite.

$$x^\alpha$$ is integrable on $$(1,\infty) \Leftrightarrow \alpha <-1$$.

Choose $$b$$ so that $$1/q-1$$.

Then $$x^{-b}\chi_{(1,\infty)} \in L^q$$ but $$\notin L^p$$ because $$x^{-bq}$$ is integrable because the exponent $$-bq<-1$$ and $$x^{-bp}$$ isn’t integrable because the exponent $$-bp>-1$$. Now I found a function that is in $$L^p$$ but not in $$L^q$$. But that doesn’t really show that $$L^p \subsetneq L^q$$, meaning $$L^p$$ is a proper subset of $$L^q$$, right (because I don’t know if every element of $$L^p$$ is also an element of $$L^q$$)?