Math 597. Homework 5 Bryden Cais

(1) (a) Let f be non-decreasing, right continuous function. Set

S_{1} = {∑(f(b_{i}) − f(a_{i})) : E ⊂ ⋃_{i}(a_{i}, b_{i}]} ,

S_{2} = {∑(β_{i} − α_{i}) : f(E) ⊂ ⋃_{i}[α_{i}, β_{i}]} ,

and recall that μ_{f}^{∗(E)} = inf S_{1} and μ_{leb}^{∗}(f(E)) = inf S_{2} where in each case the infimum is taken over all coverings by intervals of the sets E, f(E) respectively (note that it is inconsequential that we are taking closed coverings for the Lebesgue measure). Now since f is non-decreasing, if E ⊂ ⋃_{i}(a_{i}, b_{i}] then f(E) ⊂ ⋃_{i}[f(a_{i}), f(b_{i})] (since for any interval I_{i} = (a_{i}, b_{i}] we have f(a_{i}) ≤ f(x) ≤ f(b_{i}) for all x ∈ I_{i}). It follows that every covering by intervals (a_{i}, b_{i}] gives a covering [f(a_{i}), f(b_{i})] of f(E) so that every element of S_{1} is an element of S_{2}. It follows that inf S_{1} ≥ inf S_{2} or

μ_{f}^{∗}(E) ≥ μ_{leb}^{∗}(f(E)).

(b) Let f : [0, 1] → R be given by

f(x) = \begin{cases}1 & if x = 1 \\ 0 & otherwise \end{cases} .

Observe that f is nondecreasing and right continuous. However we see that

μ_{f}^{∗}((0, 1]) = f(1) − f(0) = 1

while

μ_{leb}^{∗}(f(E)) = μ_{leb}^{∗}({0, 1}) = 0.

(2) Let f(x, y) = min{x, y} and define

λ((x_{1}, x_{2}] × (y_{1}, y_{2}]) = f(x_{2}, y_{2}) + f(x_{1}, y_{1}) − f(x_{1}, y_{2}) − f(x_{2}, y_{1}).

Let Q = {∅} ∪ {(x_{1}, x_{2}] × (y_{1}, y_{2}] : x_{1} < x_{2}, y_{1} < y_{2}}. We determine a simplified geometric description of μ_{f}^{∗}. Observe that for any rectangle R lying entirely above the line x = y we have f(x_{i}, y_{i}) = min{x_{i}, y_{i}} = x_{i} for all vertices x_{i}, y_{i} of R, and therefore

λ(R) = f(x_{2}, y_{2}) + f(x_{1}, y_{1}) − f(x_{1}, y_{2}) − f(x_{2}, y_{1}) = x_{2} + x_{1} − x_{1} − x_{2} = 0.

Similarly, for any rectangle R lying completely below the line y = x, we have via a similar computation that λ(R) = 0. Therefore, the only rectangles R having nonzero pre-measure are those that intersect the line y = x. Now if a rectangle R intersects the line y = x, we may assume that R is in fact a square and the line y = x lies on the diagonal: indeed, we may “lop” off the other portions of R that lie entirely above or below the diagonal. Now for a square R whose diagonal lies on the line y = x it is not hard to see that

λ(R) = f(x_{2}, y_{2}) + f(x_{1}, y_{1}) − f(x_{1}, y_{2}) − f(x_{2}, y_{1}) = x_{2} + y_{1} − x_{1} − y_{1} = x_{2} − x_{1},

so that λ(R) = \frac{1}{\sqrt{2}}| diam R|. It follows that μ_{f}^{∗}(E) = \frac{1}{\sqrt{2}}μ(E∩\{y = x\}) where μ is Lebesgue measure, i.e., μ_{f}^{∗}(E) gives the “1/\sqrt{2} times the amount of E that lies on the line y = x.”

(3) Fix E ⊂ [0, 1]. Let

f(x) = \begin{cases}x & if x ∈ E \\ x + 5 & ifx ∉ E \end{cases} .

Observe that

f^{−1}(α) = \begin{cases}∅ & if x ∉ E ∪ ([0, 1] \ E + 5) \\ α & if α ∈ E \\ α − 5 & if α ∈ [0, 1] \ E + 5 \end{cases} ,

so that in any case f^{−1}(α) is empty or a single point for all α ∈ R, whence f^{−1}(α) is measurable for all α ∈ R.

{(a) \atop (b)} On the other hand, f need not be measurable. Indeed, if E is not measurable then [0, 1] \ E is

not measurable and we have [0, 1]\E = {x : f(x) ≥ 5} is not measurable so f is non-measurable.

(4) Recall from class that we defined a continuous function f : [0, 1] → [0, 1] with continuous inverse that mapped a set S of measure 1 to a set of measure 0. Let E ⊂ S be any non-measurable set. Observe that f(E) ⊂ f(S) must have outer measure 0 and is hence measurable. Define

g(x) = \begin{cases}1 & if x ∈ f(E) \\ −1 & otherwise \end{cases} .

Observe that since f(E) is measurable,

{x : g(x) ≥ c}

is measurable (since it is either f(E) or [0, 1]), and hence g is measurable. However, we have

{x : g ∘ f(x) ≥ 0} = f^{−1}({y : g(y) ≥ 0}) = f^{−1}f(E) = E,

which is non-measurable (where we have used that f^{−1} is continuous and f is bijective). It follows that g ∘ f is not a measurable function.