Spring 2021.6 #real_analysis/qual/work
This problem may be much harder than expected. Recommended skip.
Let f:R×R→R be a measurable function and for x∈R define the set Ex:={y∈R | μ(z∈R | f(x,z)=f(x,y))>0}. Show that the following set is a measurable subset of R×R: E:=⋃x∈R{x}×Ex.
Hint: consider the measurable function h(x,y,z):=f(x,y)−f(x,z).
Fall 2021.4 #real_analysis/qual/completed
Let f be a measurable function on R. Show that the graph of f has measure zero in R2.
solution:
Write Γ:={(x,f(x)) | x∈R}⊆Rd. Then μ(Γ)=∫RdχΓdμ=∫Rd−1∫RχΓ(x,y)dydx=∫Rd−10dx=0, using that ∫RχΓ(x,y)dy=0 since if x is fixed then χΓ(x,y)={f(x)} is a point with measure zero. Since f is measurable, Γ is a measurable set and χΓ is measurable. Since the iterated integral was finite, the equalities are justified by Fubini-Tonelli.
Spring 2020.4 #real_analysis/qual/completed
Let f,g∈L1(R). Argue that H(x,y):=f(y)g(x−y) defines a function in L1(R2) and deduce from this fact that (f∗g)(x):=∫Rf(y)g(x−y)dy defines a function in L1(R) that satisfies ‖f∗g‖1≤‖f‖1‖g‖1.
Just do it! Sort out the justification afterward. Use Tonelli.
- Tonelli: non-negative and measurable yields measurability of slices and equality of iterated integrals
- Fubini: f(x,y)∈L1 yields integrable slices and equality of iterated integrals
- F/T: apply Tonelli to |f|; if finite, f∈L1 and apply Fubini to f
- See Folland’s Real Analysis II, p. 68 for a discussion of using Fubini and Tonelli.
solution:
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If these norms can be computed via iterated integrals, we have ‖f∗g‖1:=∫R|(f∗g)(x)|dx:=∫R|∫RH(x,y)dy|dx:=∫R|∫Rf(y)g(x−y)dy|dx≤∫R∫R|f(y)g(x−y)|dxdy:=∫R∫R|H(x,y)|dxdy:=∫R2|H|dμR2:=‖H‖L1(R2). So it suffices to show ‖H‖1<∞.
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A preliminary computation, the validity of which we will show afterward: ‖H‖1:=‖H‖L1(R2)=∫R(∫R|f(y)g(x−y)|dy)dxTonelli=∫R(∫R|f(y)g(x−y)|dx)dyTonelli=∫R(∫R|f(y)g(t)|dt)dysetting t=x−y,dt=−dx=∫R(∫R|f(y)|⋅|g(t)|dt)dy=∫R|f(y)|⋅(∫R|g(t)|dt)dy:=∫R|f(y)|⋅‖g‖1dy=‖g‖1∫R|f(y)|dythe norm is a constant:=‖g‖1‖f‖1<∞by assumption.
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We’ve used Tonelli twice: to equate the integral to the iterated integral, and to switch the order of integration, so it remains to show the hypothesis of Tonelli are fulfilled.
H is measurable on R2:
proof (?):
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It suffices to show ˜f(x,y):=f(y) and ˜g(x,y):=g(x−y) are both measurable on R2.
- Then use that products of measurable functions are measurable.
- f∈L1 by assumption, and L1 functions are measurable by definition.
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The function (x,y)↦g(x−y) is measurable on R2:
- g is measurable on R by assumption, so the cylinder function G(x,y):=g(x) on R2 is measurable (result from course).
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Define a linear transformation ` \begin{align*} T \coloneqq \begin{bmatrix} 1 & -1 \ 0 & 1 \end{bmatrix} \in \operatorname{GL}_2({\mathbf{R}}) && \implies ,,, T \begin{bmatrix} x \ y \end{bmatrix}
[x−y y] ,\end{align*} `{=html} and linear functions are measurable. - Write ˜g(x−y):=G(x−y,y):=(G∘T)(x,y), and compositions of measurable functions are measurable.
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Apply Tonelli to |H|
- H measurable implies |H| is measurable.
- |H| is non-negative.
- So the iterated integrals are equal in the extended sense
- The calculation shows the iterated integral is finite, so ∫|H| is finite and H is thus integrable on R2.
Note: Fubini is not needed, since we’re not calculating the actual integral, just showing H is integrable.
Spring 2019.4 #real_analysis/qual/completed
Let f be a non-negative function on Rn and A={(x,t)∈Rn×R:0≤t≤f(x)}.
Prove the validity of the following two statements:
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f is a Lebesgue measurable function on Rn⟺A is a Lebesgue measurable subset of Rn+1
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If f is a Lebesgue measurable function on Rn, then m(A)=∫Rnf(x)dx=∫∞0m({x∈Rn:f(x)≥t})dt
- See Stein and Shakarchi p.82 corollary 3.3.
- Tonelli
- Important trick! {(x,t) | 0≤t≤f(x)}={f(x)≥t}∩{t≥0}
solution:
proof (a, $\implies$):
⟹:
- Suppose f:Rn→R is a measurable function.
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Rewrite A:
A={(x,t)∈Rd×R | 0≤t≤f(x)}={(x,t)∈Rd×R | 0≤t<∞}∩{(x,t)∈Rd×R | t≤f(x)}=(Rd×[0,∞))∩{(x,t)∈Rd×R | f(x)−t≥0}:=(Rd×[0,∞))∩H−1([0,∞)),
where we define
H:Rd×R→R(x,t)↦f(x)−t.
- Note: this is “clearly” measurable!
- If we can show both sets are measurable, we’re done, since σ-algebras are closed under countable intersections.
- The first set is measurable since it is a Borel set in Rd+1.
- For the same reason, it suffices to show H is a measurable function.
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Define cylinder functions
F:Rd×R→R(x,t)↦f(x)
and
G:Rd×R→R(x,t)↦t
- F is a cylinder of f, and since f is measurable by assumption, F is measurable.
- G is a cylinder on the identity for R, which is measurable, so G is measurable.
- Define H:Rd→R(x,t)↦F(x,t)−G(x,t):=f(x)−t, which are linear combinations of measurable functions and thus measurable.
proof (a, $\impliedby$):
⟸:
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Suppose A is a measurable set.
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A corollary of Tonelli applied to χX: if E is measurable, then for a.e. t the following slice is measurable: At:={x∈Rd | (x,t)∈A}={x∈Rd | f(x)≥t≥0}=f−1([t,∞)).
- But maybe this isn’t enough, because we need f−1([α,∞)) for all α
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But the other slice is also measurable for a.e. x: Ax:={t∈R | (x,t)∈A}={t∈R | 0≤t≤f(x)}={t∈R | t∈[0,f(x)]}=[0,f(x)].
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Moreover the function x↦m(Ax) is a measurable function of x
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Now note m(Ax)=f(x)−0=f(x), so f must be measurable.
proof (of b):
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Writing down what the slices are A={(x,t)∈Rn×R | 0≤t≤f(x)}At={x∈Rn | t≤f(x)}.
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Then ∫Rnf(x) dx=∫Rn∫f(x)01 dt dx=∫Rn∫∞0χA dt dxF.T.=∫∞0∫RnχA dx dt=∫∞0m(At) dt, where we just use that ∫∫χA=m(A)
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By Tonelli, all of these integrals are equal.
- This is justified because f was assumed measurable on Rn, thus by (a) A is a measurable set and thus χA is a measurable function on Rn×R.
Fall 2018.5 #real_analysis/qual/completed
Let f≥0 be a measurable function on R. Show that ∫Rf=∫∞0m({x:f(x)>t})dt
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Claim: If E⊆Ra×Rb is a measurable set, then for almost every y∈Rb, the slice Ey is measurable and
m(E)=∫Rbm(Ey)dy.
- Set g=χE, which is non-negative and measurable, so apply Tonelli.
- Conclude that gy=χEy is measurable, the function y↦∫gy(x)dx is measurable, and ∫∫gy(x)dxdy=∫g.
- But ∫g=m(E) and ∫∫gy(x)dxdy=∫m(Ey)dy.
solution:
Note: f is a function R→R in the original problem, but here I’ve assumed f:Rn→R.
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Since f≥0, set E:={(x,t)∈Rn×R | f(x)>t}={(x,t)∈Rn×R | 0≤t<f(x)}.
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Claim: since f is measurable, E is measurable and thus m(E) makes sense.
- Since f is measurable, F(x,t):=t−f(x) is measurable on Rn×R.
- Then write E={F<0}∩{t≥0} as an intersection of measurable sets.
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We have slices Et:={x∈Rn | (x,t)∈E}={x∈Rn | 0≤t<f(x)}Ex:={t∈R | (x,t)∈E}={t∈R | 0≤t≤f(x)}=[0,f(x)].
- Et is precisely the set that appears in the original RHS integrand.
- m(Ex)=f(x).
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Claim: χE satisfies the conditions of Tonelli, and thus m(E)=∫χE is equal to any iterated integral.
- Non-negative: clear since 0≤χE≤1
- Measurable: characteristic functions of measurable sets are measurable.
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Conclude:
- For almost every x, Ex is a measurable set, x↦m(Ex) is a measurable function, and m(E)=∫Rnm(Ex)dx
- For almost every t, Et is a measurable set, t↦m(Et) is a measurable function, and m(E)=∫Rm(Et)dt
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On one hand, m(E)=∫Rn+1χE(x,t)=∫R∫RnχE(x,t)dtdxby Tonelli=∫Rnm(Ex)dxfirst conclusion=∫Rnf(x)dx.
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On the other hand, m(E)=∫Rn+1χE(x,t)=∫R∫RnχE(x,t)dxdtby Tonelli=∫Rm(Et)dtsecond conclusion.
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Thus ∫Rnfdx=m(E)=∫Rm(Et)dt=∫Rm({x | f(x)>t}).
Fall 2015.5 #real_analysis/qual/completed
Let f,g∈L1(R) be Borel measurable.
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Show that
- The function F(x,y):=f(x−y)g(y) is Borel measurable on R2, and
- For almost every x∈R, the function f(x−y)g(y) is integrable with respect to y on R.
- Show that f∗g∈L1(R) and ‖f∗g‖1≤‖f‖1‖g‖1
solution:
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F∈B(R2):
- Write a function ˜f(x,y):=f(x)
- Write a linear transformation T=[100−1]∈GL2, so T[x,y]=[x−y,0]
- Write f(x−y):=(˜f∘T)(x,y), which is a composition of measurable functions and thus measurable.
- A product of measurable functions is measurable.
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f∗g∈L1(R): estimate ∫|f∗g|dμ=∫R∫R|f(x−y)g(y)|dxdy=∫R∫R|f(x−y)||g(y)|dxdy=∫R|g(y)|∫R|f(x−y)|dxdy=‖g‖1‖f‖1, where we’ve used translation invariance of the L1 norm and Fubini-Tonelli justified by the finite result.
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Fx(y):=f(x−y)g(y) is integrable with respect to y for almost every x:
- This follows from Fubini-Tonelli, which says that if F(x,y) is integrable, the slices Fx(y) are integrable for almost every x. Here take F(x,y):=f(x−y)g(y).
Spring 2014.5 #real_analysis/qual/work
Let f,g∈L1([0,1]) and for all x∈[0,1] define F(x):=∫x0f(y)dyandG(x):=∫x0g(y)dy.
Prove that ∫10F(x)g(x)dx=F(1)G(1)−∫10f(x)G(x)dx