Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of measurable functions defined on a subset $E$ of $\mathbb{R}^d$. Define $\sup_n f_n(x)=\sup\{f_n(x)\}$ where $x\in E$. In a proof of the statement that $\sup_n f_n(x)$ is measurable given in Stein and Shakarchi, the authors claim that $$\{x\in E\mid\sup_nf_n(x)>a\}=\bigcup_{n=1}^\infty\{x\in E\mid f_n(x)>a\}.$$

I am trying to understand why this is true. For the first inclusion, if we let $x$ be an element of the set on the left, then $\sup_nf_n(x)>a$. By definition of $\sup$, for every $\epsilon$, there exists an $f_k\in\{f_n\}$ such that $\sup_nf_n(x)-\epsilon<f_k(x)$. But then $a<f_k(x)+\epsilon$, and since $\epsilon$ was arbitrary, the result follows.

Is this correct? If not, what am I missing?


The end of your proof is incorrect. The problem is that you have proven that for each $x$ and $\varepsilon$ there is a $k$ with $a < f_k(x) + \varepsilon$. But that does not imply that there is a $k$ with $a < f_k(x)$. For example, consider the case where $a = 0$ and $f_k \equiv 0$ for all $k$. Then $a < f_k(x) + \varepsilon$ for every $\varepsilon > 0$, but there is no $k$ with $f_k(x) > a$.

Instead, you need to choose a particular value for $\varepsilon$. The trick is that because the supremum is strictly larger than $a$, there must be a $k$ such that $f_k(x)$ is squeezed between $a$ and the supremum. By choosing $\varepsilon = \sup f_n(x) - a > 0$ you obtain that there is a $k$ such that $$f_k(x) > \sup f_n(x) - a = \sup f_n(x) - (\sup f_n(x) - a) = a.$$


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