Fubini-Tonelli

Spring 2021.6 #real_analysis/qual/work

warnings:

This problem may be much harder than expected. Recommended skip.

Let f:R×RR be a measurable function and for xR define the set Ex:={yR | μ(zR | f(x,z)=f(x,y))>0}. Show that the following set is a measurable subset of R×R: E:=xR{x}×Ex.

Hint: consider the measurable function h(x,y,z):=f(x,y)f(x,z).

Fall 2021.4 #real_analysis/qual/completed

problem (?):

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)) | xR}Rd. Then μ(Γ)=RdχΓdμ=Rd1RχΓ(x,y)dydx=Rd10dx=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,gL1(R). Argue that H(x,y):=f(y)g(xy) defines a function in L1(R2) and deduce from this fact that (fg)(x):=Rf(y)g(xy)dy defines a function in L1(R) that satisfies fg1f1g1.

strategy:

Just do it! Sort out the justification afterward. Use Tonelli.

concept:

    
  • 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, fL1 and apply Fubini to f
  • See Folland’s Real Analysis II, p. 68 for a discussion of using Fubini and Tonelli.
solution:
  • If these norms can be computed via iterated integrals, we have fg1:=R|(fg)(x)|dx:=R|RH(x,y)dy|dx:=R|Rf(y)g(xy)dy|dxRR|f(y)g(xy)|dxdy:=RR|H(x,y)|dxdy:=R2|H|dμR2:=HL1(R2). So it suffices to show H1<.

  • A preliminary computation, the validity of which we will show afterward: H1:=HL1(R2)=R(R|f(y)g(xy)|dy)dxTonelli=R(R|f(y)g(xy)|dx)dyTonelli=R(R|f(y)g(t)|dt)dysetting t=xy,dt=dx=R(R|f(y)||g(t)|dt)dy=R|f(y)|(R|g(t)|dt)dy:=R|f(y)|g1dy=g1R|f(y)|dythe norm is a constant:=g1f1<by assumption.

  • 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.

claim:

H is measurable on R2:

proof (?):

    
  • It suffices to show ˜f(x,y):=f(y) and ˜g(x,y):=g(xy) are both measurable on R2.
    • Then use that products of measurable functions are measurable.
  • fL1 by assumption, and L1 functions are measurable by definition.
  • The function (x,y)g(xy) 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).
    • 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}

      [xy y] ,\end{align*} `{=html} and linear functions are measurable.
    • Write ˜g(xy):=G(xy,y):=(GT)(x,y), and compositions of measurable functions are measurable.
  • 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:0tf(x)}.

Prove the validity of the following two statements:

  • f is a Lebesgue measurable function on RnA is a Lebesgue measurable subset of Rn+1

  • If f is a Lebesgue measurable function on Rn, then m(A)=Rnf(x)dx=0m({xRn:f(x)t})dt

concept:

    
  • See Stein and Shakarchi p.82 corollary 3.3.
  • Tonelli
  • Important trick! {(x,t) | 0tf(x)}={f(x)t}{t0}
solution:
proof (a, $\implies$):

:

  • Suppose f:RnR is a measurable function.
  • Rewrite A: A={(x,t)Rd×R | 0tf(x)}={(x,t)Rd×R | 0t<}{(x,t)Rd×R | tf(x)}=(Rd×[0,)){(x,t)Rd×R | f(x)t0}:=(Rd×[0,))H1([0,)), where we define H:Rd×RR(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.
  • Define cylinder functions F:Rd×RR(x,t)f(x) and G:Rd×RR(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:RdR(x,t)F(x,t)G(x,t):=f(x)t, which are linear combinations of measurable functions and thus measurable.
proof (a, $\impliedby$):

:

  • Suppose A is a measurable set.

  • A corollary of Tonelli applied to χX: if E is measurable, then for a.e. t the following slice is measurable: At:={xRd | (x,t)A}={xRd | f(x)t0}=f1([t,)).

    • But maybe this isn’t enough, because we need f1([α,)) for all α
  • But the other slice is also measurable for a.e. x: Ax:={tR | (x,t)A}={tR | 0tf(x)}={tR | t[0,f(x)]}=[0,f(x)].

  • Moreover the function xm(Ax) is a measurable function of x

  • Now note m(Ax)=f(x)0=f(x), so f must be measurable.

proof (of b):

    
  • Writing down what the slices are A={(x,t)Rn×R | 0tf(x)}At={xRn | tf(x)}.

  • Then Rnf(x) dx=Rnf(x)01 dt dx=Rn0χA dt dxF.T.=0RnχA dx dt=0m(At) dt, where we just use that χA=m(A)

  • 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 f0 be a measurable function on R. Show that Rf=0m({x:f(x)>t})dt

concept:

    
  • Claim: If ERa×Rb is a measurable set, then for almost every yRb, 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 ygy(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 RR in the original problem, but here I’ve assumed f:RnR.

  • Since f0, set E:={(x,t)Rn×R | f(x)>t}={(x,t)Rn×R | 0t<f(x)}.

  • Claim: since f is measurable, E is measurable and thus m(E) makes sense.

    • Since f is measurable, F(x,t):=tf(x) is measurable on Rn×R.
    • Then write E={F<0}{t0} as an intersection of measurable sets.
  • We have slices Et:={xRn | (x,t)E}={xRn | 0t<f(x)}Ex:={tR | (x,t)E}={tR | 0tf(x)}=[0,f(x)].

    • Et is precisely the set that appears in the original RHS integrand.
    • m(Ex)=f(x).
  • Claim: χE satisfies the conditions of Tonelli, and thus m(E)=χE is equal to any iterated integral.

    • Non-negative: clear since 0χE1
    • Measurable: characteristic functions of measurable sets are measurable.
  • Conclude:

    • For almost every x, Ex is a measurable set, xm(Ex) is a measurable function, and m(E)=Rnm(Ex)dx
    • For almost every t, Et is a measurable set, tm(Et) is a measurable function, and m(E)=Rm(Et)dt
  • On one hand, m(E)=Rn+1χE(x,t)=RRnχE(x,t)dtdxby Tonelli=Rnm(Ex)dxfirst conclusion=Rnf(x)dx.

  • On the other hand, m(E)=Rn+1χE(x,t)=RRnχE(x,t)dxdtby Tonelli=Rm(Et)dtsecond conclusion.

  • Thus Rnfdx=m(E)=Rm(Et)dt=Rm({x | f(x)>t}).

Fall 2015.5 #real_analysis/qual/completed

problem (?):

Let f,gL1(R) be Borel measurable.

  • Show that
    • The function F(x,y):=f(xy)g(y) is Borel measurable on R2, and
    • For almost every xR, the function f(xy)g(y) is integrable with respect to y on R.
  • Show that fgL1(R) and fg1f1g1
solution:

    
  • FB(R2):

    • Write a function ˜f(x,y):=f(x)
    • Write a linear transformation T=[1001]GL2, so T[x,y]=[xy,0]
    • Write f(xy):=(˜fT)(x,y), which is a composition of measurable functions and thus measurable.
    • A product of measurable functions is measurable.
  • fgL1(R): estimate |fg|dμ=RR|f(xy)g(y)|dxdy=RR|f(xy)||g(y)|dxdy=R|g(y)|R|f(xy)|dxdy=g1f1, where we’ve used translation invariance of the L1 norm and Fubini-Tonelli justified by the finite result.

  • Fx(y):=f(xy)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(xy)g(y).

Spring 2014.5 #real_analysis/qual/work

problem (?):

Let f,gL1([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

#real_analysis/qual/work #real_analysis/qual/completed