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WA 4: Solutions
- Let the function f on [0, 1] be defined as
f (x) =
q if^ x^ =^
p q where^ p^ and^ q^ are coprime integers,^0 ≤^ p^ ≤^ q, 0 if x is irrational or x=0.
(a) Prove that f is continuous at every irrational point and at 0, and that f is discontinuous at every rational point of the interval (0, 1];
(b) Is f Riemann integrable? If so, find R
∫^1
0
f (x) dx.
(c) Is f Lebesgue integrable? If so, find
[0,1]
f.
Solution. (a) Let x 0 be irrational or 0. Given > 0, there is only a finite number of proper rational fractions p/q such that 1/q ≥ (those are fractions with q ≤ 1 / and 0 ≤ p < q). Let δ be the distance from x 0 to the closest rational number with this property. Then if x = p/q is such that |x−x 0 | < δ then |f (x)−f (x 0 )| = 1/q < , and if x is irrational and |x − x 0 | < δ then |f (x) − f (x 0 )| = 0 < . Thus, f is continuous at x 0. If x 0 = p/q ∈ (0, 1] then there are irrational numbers x arbitrary close to x 0 , and |f (x) − f (x 0 )| = 1/q > 0. Therefore, f is discontinuous at x 0. (b) Since the set of discontinuities of f coincides with the set of rational num- bers in (0, 1], and thus countable, and any countable set has measure zero, by the Lebesgue theorem f is Riemann integrable. Since
R
∫^1
0
f (x) dx = R
0
f (x) dx = sup s,
where s are lower Darboux sums, and for any partition of [0, 1] there are irrational
points in each subinterval, we have s = 0 for all s, and therefore R
∫^1
0
f (x) dx = 0.
(c) Since f is Riemann integrable, f is also Lebesque integrable, and
∫
[0,1]
f = R
∫^1
0
f (x) dx = 0.
- Show that if f is measurable real-valued function and g is a continuous function defined on (−∞, ∞) then the composition g ◦ f is measurable.
Solution. For any extended real number α, the set
{x : g(x) > α} = g−^1 (α, ∞)
is open (as a pre-image of an open set under a continuous mapping), and therefore can be represented as a finite or countable union of disjoint intervals (an, bn). Since the set
{x : an < f (x) < bn} = f −^1 (an, bn) 1
is measurable for any n, the set
{x : g(f (x)) > α} =
n
f −^1 (an, bn)
is measurable. Therefore, g ◦ f is a measurable function.
- An extended real-valued function f is called lower semicontinuous at the point y if f (y) 6 = −∞ and f (y) ≤ lim x→y
f (x). Similarly, f is called upper semicon-
tinuous at the point y if f (y) 6 = +∞ and f (y) ≥ lim x→y f (x). We say that f is lower
(upper) semicontinuous on an interval if it is lower (upper) semicontinuous at each point of the interval. Clearly, f is upper semicontinuos if and only if −f is lower semicontinuous. (a) Let f (y) be finite. Prove that f is upper semicontinuous at y if and only if given > 0, there is a δ > 0 such that f (y) ≥ f (x) − for all x with |x − y| < δ. (b) A function f is continuous (at a point of in an interval) if and only if it is both upper and lower semicontinuous (at a point of in an interval). (c) A step function φ on [a, b] is upper semicontinuous if and only if for a partition a = x 0 < x 1 < · · · < xn = b defining φ and for each i = 0,... , n the value φ(xi) is greater than or equal to the maximal of the two values assumed on (xi− 1 , xi) and (xi, xi+1) (for the endpoints x 0 = a and xn = b, one of the two subintervals is void). (d) A function f on [a, b] is upper semicontinuous if and only if there is a mono- tone decreasing sequence {φn} of upper semicontinuous step functions on [a, b] such that for each x ∈ [a, b], f (x) = lim n φn(x).
Solution. (a) follows from the equivalence of two definitions of a lower (resp., upper) limit. (b) f is continuous at y if and only if
f (y) = lim x→y f (x) = lim x→y f (x) = lim x→y f (x).
This implies that f is both lower semicontinuous and upper semicontinuous at y. Conbversely, if f is both lower semicontinuous and upper semicontinuous at y then
lim x→y f (x) ≤ f (y) ≤ lim x→y f (x) ≤ lim x→y f (x).
Therefore, we have all equalities in this chain of inequalities, and f is continuous at y. (c) If φ(x) = ci− 1 for x ∈ (xi− 1 , xi) and φ(x) = ci for x ∈ (xi, xi+1) then, clearly,
xlim→x i φ(x) = max{ci− 1 , ci}. Therefore, φ is upper semicontinuous at xi if and only
if φ(xi) ≥ max{ci− 1 , ci}. If y is an interior point of any subinterval (xi, xi+1) in the partition for φ then φ is continuous at y, and hence, by part (b) φ is always upper semicontinuous at such a y.
at y then, by the result of Problem 3(a), given > 0 there exists a δ > 0 such that f (y) ≥ sup |x−y|<δ
f (x) − . Therefore,
f (y) ≥ inf δ> 0
sup |x−y|<δ
f (x) − = h(y) − .
Since > 0 was arbitrary, f (y) = h(y). The statement for g is proved analogously. (b) Since f is bounded, h(y) 6 = ∞ for every y ∈ [a, b]. Given such y, for every δ > 0 one has
lim x→y h(x) ≤ lim x→y sup |t−x|<δ
f (t) ≤ sup |x−y|<δ
sup |t−x|<δ
f (t) = sup |t−y|< 2 δ
f (t).
Therefore, lim x→y h(x) ≤ inf δ> 0 sup |t−y|< 2 δ
f (t) = h(y),
i.e., h is upper semicontinuous. The statement for g is proved analogously.
- (a) Let f be bounded function on [a, b], and let h be its upper envelope. Then
R
∫ (^) b a f^ (x)dx^ =^
∫ (^) b a h. Hint: Use the following line of reasoning, and explain the details. If φ ≥ f is a step function, then φ ≥ h except at a finite number of points, and so
∫ (^) b a h^ ≤ R
∫ (^) b a f^ (x)dx. But there is a sequence^ {φn}^ of step functions such that^ φn^ ↓^ h^ (see Problem 3). By the Bounded Convergence Theorem, we have
∫ (^) b a h^ = lim^
∫ (^) b a φn^ ≥ R
∫ (^) b a f^ (x)dx. (b) Use part (a) to prove the Lebesque theorem: a bounded function f on [a, b] is Riemann integrable if and only if the set of points at which f is discontinuous has measure zero.
Solution. (a) If φ ≥ f is a step function then h(y) = inf δ> 0 sup |x−y|<δ
f (x) ≤ inf δ> 0 sup |x−y|<δ
φ(x) = φ(y)
except, maybe, when y is a partition point for the step function φ. Thus, φ ≥ h except at a finite number of points. We know from Problem 3(d) that there is a sequence {φn} of step functions such that φn ↓ h. Then h is measurable as a limit of measurable functions. Since φ 1 ≥ φn ≥ h for every n, we have |φn| ≤ |φ 1 | + |h|. Since |h| is bounded by the same constant as |f |, |h| is Lebesque integrable, and we also have that |φ 1 | is Lebesgue integrable. Thus, |φ 1 | + |h| is Lebesgue integrable. Therefore, we can apply the Lebesgue convergence theorem, and get
∫ (^) b
a
h = lim n→∞
∫ (^) b
a
φn = lim n→∞
R
∫ (^) b
a
φn(x) dx
= inf φ≥h
R
∫ (^) b
a
φ(x) dx
= inf φ≥f
R
∫ (^) b
a
φ(x) dx
= R
∫ (^) b
a
f (x)dx.
(b) The analogue of part (a) for the lower envelope g of f can be established analogously. Then ∫ (^) b
a
g = R
∫ (^) b
a
f (x)dx ≤ R
∫ (^) b
a
f (x)dx =
∫ (^) b
a
h.
The function f is Riemann integrable if and only if this inequality is an equality. This, in turn, happens if and only if
∫ (^) b a (h^ −^ g) = 0. Since^ h^ −^ g^ ≥^ 0 this is possible if and only if h = g almost everywhere. By Problem 4(a), this, in turn, means that f is continuous at all points of [a, b] except on the set of measure zero.
