Past exam of the martingales and financial mathematics course of the University of Oxford / 2008 / A / 4

A Borel measurable function has $f^{-1}(B)\in \mathcal{B}$ for every Borel set $B$. Thus the upper-ray condition is necessary.

For sufficiency, note that the rays generate the stipulated sigma-algebra: $(a,b]=(a,\infty )\setminus (b,\infty )$, and $(y,\infty )={\bigcup }_{m>y}(y,m]$, with integer $m$. The familyis a sigma-algebra because inverse images preserve complements and countable unions. If it contains all upper rays, it contains their generated Borel sigma-algebra. Therefore upper-ray preimages suffice to prove Borel measurability.

Every open interval is Borel, since $(a,b)={\bigcup }_{m\ge 1}(a,b-1/m]$, omitting empty pieces. Open subsets of the line are countable unions of open intervals and hence Borel. Conversely, each half-open interval is obtained from open sets by sigma-algebra operations, so this is the usual Borel sigma-algebra.

A continuous function has open, and therefore Borel, inverse images of open sets. Since open sets generate the target Borel sigma-algebra, it is a Borel measurable function. A discontinuous example is $1_{[0,\infty )}$, whose only level sets are built from the Borel half-lines; it jumps at zero.

For a lower semicontinuous function, each upper-ray inverse image is open, hence Borel, so part (a)(i) applies directly. For an upper semicontinuous $h$, the function $-h$ is lower semicontinuous becauseis open. Negation is continuous and Borel measurable, so $h=-(-h)$ is Borel measurable too. Thus both semicontinuity classes are Borel measurable.

Boundedness of $f$ makes $\phi$ finite and nonnegative. Suppose $\phi (x_0)<a$. By its infimum definition, some $n$ satisfies ${\phi }_n(x_0)<a$. If $|x-x_0|<1/(2n)$, thenThe supremum over the smaller interval cannot exceed that over the larger one. HenceThus $\{\phi <a\}$ is open for every $a$, proving $\phi$ is upper semicontinuous. Measurability of the original $f$ was never needed: these are purely pointwise oscillation bounds.

If $f$ is continuous at $x_0$, choose a small interval on which $|f(s)-f(x_0)|<ϵ/2$. The triangle inequality makes its oscillation less than $ϵ$, so $\phi (x_0)=0$.

Conversely, if $\phi (x_0)=0$, then for every $ϵ>0$ some $n$ has ${\phi }_n(x_0)<ϵ$. As $x_0$ itself belongs to the interval used in that supremum, $|f(x)-f(x_0)|<ϵ$ for $|x-x_0|<1/n$. Thus $f$ is continuous at $x_0$.

The continuity set of a function is thereforeEach set in the intersection is open by part (i), so this is a G-delta set and in particular a Borel set, even if $f$ itself is not measurable.