2 Answers 2

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If $x \in \{f \leq a\}$, you have that $f(x) \leq a$. So $f(x) \leq a + \frac{1}{k}$ for all $k$. It has been shown that $\{f \leq a\} \subset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$.

If $x \in \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$. Then $f(x) < a + \frac{1}{k}$ for all $k$. This can only happen if $f(x) \leq a$. So $\{f \leq a\} \supset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$

The first equality of sets has been shown.

The second is similar.

If $x \in \{f < a\}$, then $a - f(x) > 0$. Choose a $k$ such that $\frac{1}{k} \leq a - f(x)$. Then $f(x) \leq a - \frac{1}{k}$. Hence $x \in \{f \leq a - \frac{1}{k} \}$. Thus, $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$.

Suppose $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$. This means that $f(x) \leq a - \frac{1}{k}$ for some $k$. So $f(x) < a$. $x \in \{f < a\}$.

Equality of the two sets has been shown.

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    I think there are a couple typos in proving the second set equality. You wrote some < signs when you needed $\leq$ to signs.2012-08-24
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The second equation can also be proved by using the first one: Let $g = -f$. Then the first equation applied to $g$ and $-a$ is

$\{g \le -a\} = \bigcap_{k=1}^\infty \left\{g < -a + \frac 1k\right\}.$

Take the complement on both sides:

$\{g > -a\} = \bigcup_{k=1}^\infty \left\{g \ge -a + \frac 1k\right\}.$

Substitute $f$ in for $-g$:

$\{f < a\} = \bigcup_{k=1}^\infty \left\{f \le a - \frac 1k\right\}.$