I have a question that is WAY over my head, and I wanted to ask any kind of help.
Suppose $f_i(x)= 1, · · · , n$ are PDFs. That is, $f_i(x)≥0$ and $\int f_i(x)dx=1$ for all $i=1, · · · , n$. Consider a function $$g(x)=\sum_{i=1}^n p_i f_i(x)$$ where $p_i > 0$ and $\sum_{i=1}^n p_i =1$ .
Show that $g(x)$ is also a PDF.
Since you have explicitly written what it takes for a function to be a PDF you now have to verify that $g(x)$ satisfies these conditions.
For $g(x)\geq 0$ you should simply note that each $p_if_i(x)$ is a product of a positive ($p_i$) factor and a non-negative function $f_i(x)$, so each $p_if_i(x)\geq0$ and of course so is their sum.
Now, $\int g(x)dx=\int\sum p_if_i(x)=\sum p_i\int f_i(x)=\sum p_i=1$, as required, where the order of summation and integration can be changed because the sum is finite.
