Show that if $f: [0,1]\to \mathbb{R}$ is lower semi-continuous then attains its minimum on $[0,1]$ Let $f: [0,1]\to \mathbb{R}$ be a lower semi-continuous function, then
$$ \liminf_{x\to a} f(x) \geq f(a), \forall a \in [0,1]$$
I have to prove that $f$ attains its minimum on $[0,1]$, that is:
$\exists x_0 \in [0,1]$ such that $f(x_0) \le f(x)$, $\forall x \in [0,1]$.
This is a problem from a past qualifying exam in Measure Theory. I'm trying to solve it but I do not know how I should begin with this problem. I did not understand where the Inf is taken. What means that limit?
 A: Here is a simple solution. 
Let $\{x_n\}\in [0,1]$ be a minimizing sequence, that is, such that 
$$\lim_{n\to \infty} f(x_n)= f^*= \inf_{x\in [0,1]}f(x).$$ Such a sequence always exists by the definition of infimum. Since $[0,1]$ is compact, the sequence $\{x_n\}$ has a convergent subsequence. Without loss of generality assume that $x_n\to \bar{x}\in [0,1].$ Then 
$$f(\bar{x})\leq \liminf_{n\to \infty}f(x_n)= \lim_{n\to \infty} f(x_n)= f^*.$$ Since $\bar{x} \in [0,1]$ and by the definition of $f^*,$ we can only have 
$$f(\bar{x})=f^*,$$ as desired.
A: The usual proof for continuous functions can be adapted easily to this case. Assume on the contrary that $f$ is unbounded below on $[0,1]$. Then it must be unbounded below on either the left or the right half of the interval $[0,1]$. Select one of the halves where $f$ is unbounded below. And continue this process to get a chain of nested intervals where $f$ is unbounded below. By nested interval principle there is a unique point $c$ which lies in all these nested intervals. Since $f$ is lower semi-continuous at $c$ there is a neighborhood of $c$ in which $f(x) \geq f(c) - 1$. And one of the nested intervals is contained in this neighborhood so that $f$ is bounded below in this nested interval. The contradiction proves that $f$ is bounded below in $[0,1]$. 
Next we apply the same nested interval principle to prove that $f$ attains its minimum value on $[0,1]$. Let $m$ be the greatest lower bound (glb) of $f$ on $[0,1]$. Then $m$ is the glb of $f$ on at least one of the two halves of $[0,1]$. Select the half for which $m$ is the glb and repeat the process to generated a sequence of nested intervals such that $m$ is the glb of $f$ on each interval in this sequence. Like before there is a unique point $c$ which lies in all the intervals of this sequence. We prove that $f(c) =m$. On the contrary let's assume that $f(c) >m$ and choose $\epsilon=(f(c) - m) /2>0$. Then by semi-continuous of $f$ at $c$ we have a neighborhood of $c$ in which $$f(x) \geq f(c) - \epsilon=\frac{f(c) +m} {2}>m$$ One of the intervals of the sequence is contained in this neighborhood and by the above inequality the glb of $f$ on this interval is greater than or equal to $(f(c) +m) /2$ and hence greater than $m$. This is a contradiction (because the glb is $m$ on each interval of the sequence). Thus $f(c) =m$ and we are done.

Both the proofs above are based on nested interval principle. It is a fun exercise to prove these properties using various other equivalent formulations of completeness of real numbers (like Dedekind's theorem, Bolzano Weierstrass, Heine Borel Theorem, least upper bound principle). Proving the properties of continuous (semi-continuous) functions on closed intervals via completeness of reals is a key to proper understanding of completeness of real numbers. 
