$m(u)=\sup\{a\in\mathbb{R}:u\geq a\text{ almost everywhere}\}$. Show that $\inf\{:||f||_2=1\}=m(u)$ Let $\mu$ be $\sigma$-finite borel measure on $\mathbb{R}$, and $u:\mathbb{R}\to\mathbb{R}$ bounded borelian function. Let's define $m(u)=\sup\{a\in\mathbb{R}:u\geq a\text{ almost everywhere}\}$. Show for linear operator $M_u:L_2[0,1]\to L_2[0,1],M_uf=uf$ that
$$
\inf\{\langle M_uf,f \rangle:||f||_2=1\}=m(u)
$$
where $\langle f,g \rangle=\int_{[0,1]}f(t)g(t)dt$
Hint says to use fact that from $\sigma$-finiteness of measure $\mu$ follows that for every set such that $\mu(B)>0$, there exists $A\subseteq B$ such that $0<\mu(A)<\infty$.
However, I do not know how to use this hint.
 A: First let's do the easier lower bound. It is straightforward to see that we in fact have that $u \geq m(u)$ almost everywhere (since $u \geq m(u) - \frac1n$ a.e. for any $n \in \mathbb{N}$ and the countable union of null sets is null). Hence if $\|f\|_2 = 1$ we have 
$$\langle M_u f, f \rangle = \int u |f|^2 d\mu \geq m(u) \int |f|^2 d \mu = m(u)$$
so $A := \inf\{\langle M_uf,f \rangle:||f||_2=1\} \geq m(u)$. Now, the upper bound is a little bit harder. Here is where we will use the hint.
By the definition of $m(u)$, for any $\varepsilon > 0$ there is a set $B_\varepsilon$ such that $u < m(u) + \varepsilon$ on $B_\varepsilon$ and such that $\mu(B_\varepsilon) > 0$. We would like to define $f = \mu(B_\varepsilon)^{-1} 1_{B_\varepsilon}$ so that $$\langle M_u f, f \rangle \leq (m(u) + \varepsilon) \int |f|^2 d \mu = m(u) + \varepsilon.$$
Unfortunately, this runs into trouble since we could have $\mu(B_\varepsilon) = \infty$. But as the hint says, we can find $C_\varepsilon \subseteq B_\varepsilon$ such that $0 < \mu(C_\varepsilon) < \infty$. Then if $f = \mu(C_\varepsilon)^{-1} 1_{C_\varepsilon}$ we have $\|f\|_2 = 1$ and $\langle M_uf, f \rangle \leq m(u) + \varepsilon$ since $u < m(u) + \varepsilon$ on $C_\varepsilon$. Hence $A \leq m(u) + \varepsilon$ and since $\varepsilon$ was arbitrary we must have $A \leq m(u)$.
Combining the two parts then gives $A = m(u)$ as desired. 
