Inclusion-exclusion-like fractional sum is positive?

Here is an analytic proof inspired by the one in the answer to question #1343375 (thanks to kubek789 for the link!).

We fix a positive integer $n$. We let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $.

We first show an auxiliary result:

Theorem 1. Let $A_{1},A_{2},\ldots,A_{n}$ be $n$ finite nonempty sets. Let $x\in\left( 0,1\right) $. Then, \begin{equation} 0<\sum\limits_{I\subseteq\left[ n\right] }\left( -1\right) ^{\left\vert I\right\vert }x^{\left\vert \bigcup_{i\in I}A_{i}\right\vert }<1 . \end{equation}

Proof of Theorem 1. Let $A$ be the set $A_{1}\cup A_{2}\cup\cdots\cup A_{n} $. Let $X$ be the set of all maps $A\rightarrow\left\{ 0,1\right\} $. We make $X$ into a probability space by assigning to a map $f:A\rightarrow \left\{ 0,1\right\} $ the probability $x^{\left\vert f^{-1}\left( 1\right) \right\vert }\left( 1-x\right) ^{\left\vert f^{-1}\left( 0\right) \right\vert }$. This probability space $X$ is precisely the $\left\vert A\right\vert $-th Cartesian power of the probability space $\left\{ 0,1\right\} $ where $1$ has probability $x$ and $0$ has probability $1-x$. In simpler terms, randomly sampling a point in $X$ is tantamount to constructing a map $A\rightarrow\left\{ 0,1\right\} $ by flipping an $x$-coin for every $a\in A$ (independently), and letting the map $A\rightarrow\left\{ 0,1\right\} $ send $a$ to $1$ if the $x$-coin has brought up heads and to $0$ if it has brought up tails. Here, an "$x$-coin" means a biased coin which brings up heads with probability $x$.

Now, let $Y\subseteq X$ be the set of all maps $f:A\rightarrow\left\{ 0,1\right\} $ such that every $i\in\left\{ 1,2,\ldots,n\right\} $ satisfies $0\in f\left( A_{i}\right) $. Then, $Y$ is the set of all maps $f:A\rightarrow\left\{ 0,1\right\} $ such that no $i\in\left\{ 1,2,\ldots,n\right\} $ satisfies $A_{i}\subseteq f^{-1}\left( 1\right) $. Thus, the probability of $Y$ (as an event in the probability space $X$) can be computed by the principle of inclusion and exclusion to be $\sum\limits _{I\subseteq\left[ n\right] }\left( -1\right) ^{\left\vert I\right\vert }x^{\left\vert \bigcup_{i\in I}A_{i}\right\vert }$. But this probability is $>0$ (since the "constant-$0$" map belongs to $Y$ and has nonzero probability) and $<1$ (since the "constant-$1$" map does not belong to $Y$, and still has nonzero probability). Thus, Theorem 1 follows.

(Please correct all stochastics terminology that I have butchered. I have not used probability spaces since I passed probability 101 some 9 years ago.)

Theorem 2. Let $A_{1},A_{2},\ldots,A_{n}$ be $n$ finite nonempty sets. Then, \begin{equation} \sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert -1}\dfrac{1}{\left\vert \bigcup_{i\in I}A_{i}\right\vert }>0 . \end{equation}

Proof of Theorem 2. Theorem 1 shows that every $x\in\left( 0,1\right) $ satisfies

$1>\sum\limits_{I\subseteq\left[ n\right] }\left( -1\right) ^{\left\vert I\right\vert }x^{\left\vert \bigcup_{i\in I}A_{i}\right\vert }$

$=1+\sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert }x^{\left\vert \bigcup_{i\in I} A_{i}\right\vert }$ (here, we have split the $I=\varnothing$ term out of the sum)

$=1-\sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert -1}x^{\left\vert \bigcup_{i\in I} A_{i}\right\vert }$.

In other words, every $x\in\left( 0,1\right) $ satisfies

$\sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert -1}x^{\left\vert \bigcup_{i\in I} A_{i}\right\vert }>0$.

We denote the left hand side of this inequality by $f\left( x\right) $. Then, we thus have shown that every $x\in\left( 0,1\right) $ satisfies $f\left( x\right) >0$. Hence, $\int_{0}^{1}\dfrac{f\left( x\right) } {x}dx>0$. But the definition of $f$ (and the rule that $\int_{0}^{1} \dfrac{x^{m}}{x}dx=\dfrac{1}{m}$ for every $m\geq1$) yields $\int_{0} ^{1}\dfrac{f\left( x\right) }{x}dx=\sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert -1}\dfrac{1}{\left\vert \bigcup_{i\in I}A_{i}\right\vert }$. Hence, $\sum\limits_{\substack{I\subseteq\left[ n\right] ;\\I\neq\varnothing}}\left( -1\right) ^{\left\vert I\right\vert -1}\dfrac{1}{\left\vert \bigcup_{i\in I}A_{i}\right\vert }>0$, and Theorem 2 is proven.