Problem about sequence of non-negative function

Question

Suppose $\{f_n\}_{n\geq 1}$ is a sequence of non-negative functions.

define $g_n = \max_{1\leq i \leq n} f_i$

I tried to show that $$ \int_{g_n \geq a} g_n d\mu \leq \sum_{i=1}^n \int_{f_i \geq a} f_i d\mu $$

It is easy to show for $n=2$

But how can show for general n?

Thanks in advance

Answer 1

Maybe induction, you know how to make the case $ n = 1 $ and $ n = 2 $. Let's assume the case $ n = k $ and prove the case $ n = k + 1 $. We have two options, or max $\max_{1 \leq i \leq {k+1}}f_i$ is $f_{k+1}$ and then the statement is proven because $$ \int_{g_{k+1} \geq a} g_{k+1} d\mu = \int_{f_{k+1} \geq a} f_{k+1} d\mu \leq \sum_{i = 1}^{k+1} \int_{f_i \geq a}f_id\mu $$ since $ \sum_{i = 1}^{k} \int_{f_i \geq a}f_i d\mu \geq 0 $, or $\max_{1 \leq i \leq {k+1}}f_i = \max_{1 \leq i \leq {k}}f_i$. In this case, $$ \int_{g_{k+1} \geq a} g_{k+1} d\mu = \int_{g_{k} \geq a} (\max_{1 \leq i \leq {k}}f_i)d\mu \leq \int_{g_{k} \geq a} (\max_{1 \leq i \leq {k}}f_i)d\mu + \int_{f_{k+1} \geq a}f_{k+1}d\mu, $$ since $f_n \geq 0, \forall n$ . Using the induction hipotesis $$ \int_{g_{k+1} \geq a} g_{k+1} d\mu \leq \sum_{i = 1}^{k+1} \int_{f_i \geq a}f_id\mu .$$

Answer 2

If $g_n(x)>a,$ then there exists $i$ such that $g_n(x)=f_i(x).$ Clearly for this $i$ and $x,$$f_i(x) = f_i(x)\chi_{f_i >a}.$ From this it follows that

$$g_n(x)\chi_{g_n>a}(x) \le \sum_{i=1}^{n}f_i(x)\chi_{f_i >a}(x).$$

Integrate over the measure space to get the result.