I'm working through Analysis on Manifolds by Munkres on my own. He proves a version of Fubini's theorem for multivariable Riemann integrals that goes like:

If $Q= A \times B \subset \mathbb{R}^n \times \mathbb{R}^m$ is a closed bounded rectangle and the Riemann integral $\int_Q f$ exists then

$$\int_Q f = \int_A \underline{\int}_B f(x,y) dy dx = \int_A \overline{\int}_B f(x,y) dy dx $$

The lower and upper integrals are needed here when, fixing $x$, $f(x,\cdot):B \to \mathbb{R}$ is bounded but not Riemann integrable over $B$. So if $f(x, \cdot)$ is integrable then we get Fubini's theorem where

$$\int_Q f = \int_A \int_B f(x,y) dy dx = \int_B \int_A f(x,y) dy dx $$

I think this could happen if $f$ is not absolutely integrable, otherwise Fubini's theorem for Lebesgue integrals applies.

My question is what are some examples where $\int_Q f$ exists but $\int_Bf(x,y)dy$ does not? More than one is welcome.


To clarify, I mean I am looking for examples where $\int_Bf(x,y) dy$ is not Riemann integrable almost everywhere over $A$, but still $f$ is Riemann integrable over $Q$, so that

$$\int_Q f \neq \int_A \int_B f(x,y) dy dx$$

  • $\begingroup$ Could you provide the theorem number of page number in the book? $\endgroup$ Commented Jan 23, 2018 at 23:14
  • $\begingroup$ @Mathemagical: Theorem 12.2 on page 100 $\endgroup$
    – AlRacoon
    Commented Jan 23, 2018 at 23:40
  • $\begingroup$ The function $F(x) = \int_B f(x,y) \, dy$ either is or is not Riemann integrable over the rectangle $A$. It really makes no sense to talk about being non-integrable for almost every $x \in A$. $\endgroup$
    – RRL
    Commented Feb 6, 2018 at 19:20
  • $\begingroup$ Have a look here $\endgroup$
    – Christoph
    Commented May 6, 2022 at 8:24

2 Answers 2


If $\int_Q f$ exists, then by Fubini's theorem for Riemann integration we have

$$\int_A \left(\overline{\int}_B f(x,y) \, dy \right) \, dx = \int_A \left(\underline{\int}_B f(x,y) \, dy \right) \, dx,$$

and, hence,

$$\int_A \left(\overline{\int}_B f(x,y) \, dy - \underline{\int}_B f(x,y) \, dy \right) \, dx = 0.$$

Since the diferrence of the upper and lower Darboux integrals is nonnegative, it follows that for almost every $x \in A$,

$$\overline{\int}_B f(x,y) \, dy = \underline{\int}_B f(x,y) \, dy ,$$

and the function $f(x, \cdot) :y \mapsto f(x,y)$ is Riemann integrable for almost every $x \in A$.

It is not difficult to find examples where $F(x) = \int_B f(x,y) \, dy$ fails to exist as a Riemann integral on some subset of $E\subset A$ of measure $0$. At points where $f(x, \cdot)$ is integrable, define $F(x) = \int_B f(x,y) \,dy$.

Technically, it is meaningless at this juncture to ask if $F$ is or is not Riemann integrable over $A$, without fully defining $F:A \to \mathbb{R}.$ The question should be is it possible to assign values to $F(x)$ for $x \in E$ such that $f$ is Riemann integrable. More specifically you are looking for an example where regardless of how those values are defined, $F$ is not Riemann integrable.

It turns out that as long as we make the assignment for $x \in E$ such that

$$\underline{\int}_B f(x,y) \, dy \leqslant F(x) \leqslant \overline{\int}_B f(x,y) \, dy ,$$

then $F$ will be Riemann integrable over $A$ with $\int_Q f = \int_A F(x) \, dx$.

To prove this we can show that since $F$ is bounded, upper and lower Darboux sums exist such that for any partition $P = P_A \times P_B$ of $Q = A \times B$ we have

$$L(P,f) \leqslant L(P_A,F) \leqslant U(P_A,F) \leqslant U(P,f).$$

Since $f$ is Riemann integrable over $Q$, for any $\epsilon > 0$ there exists a partition $P$ such that

$$U(P_A,F) - L(P_A,F) \leqslant U(P,f) - L(P,f) < \epsilon,$$

and it follows by the Riemann criterion that $F$ is integrable over $A$.

  • $\begingroup$ Thanks. It makes sense. What is the justification for $L(P,f) \leq L(P_A,F) \leq ...$? From that I do see that $\int_AF(x)dx$ exists though. $\endgroup$
    – AlRacoon
    Commented Feb 6, 2018 at 23:40
  • $\begingroup$ @AlRacoon: SInce $F(x)$ either equals the lower and upper integrals (when they are equal) or falls in between, we first show that $L(P,f) \leqslant L(P_A, \underline{\int}_B f(x,y) \, dy) \leqslant U(P_A, \underline{\int}_B f(x,y) \, dy) \leqslant U(P,f)$ using straightforward estimates and then it follows. This is shown in detail here. $\endgroup$
    – RRL
    Commented Feb 7, 2018 at 0:36

I think even a single example would make things quite clear already.

Let $A=B=[0,1]$ and define $f(x,y)$ to be 1 if $x=\frac{1}{2}, y \in \Bbb Q$ but 0 elsewhere.

$$\underline{\int}_B f(x,\frac{1}{2}) dy=0, \overline{\int}_B f(x,\frac{1}{2}) dy=1$$

And $\int_B f(x,\frac{1}{2}) dy$ does not exist.

However $\int_Q f$ does exist and $$\int_Q f = \int_A \underline{\int}_B f(x,y) dy dx = \int_A \overline{\int}_B f(x,y) dy dx =0$$

  • $\begingroup$ Thanks. I think you mean $f(\frac{1}{2},y)$ above. While this is interesting it is not really what I'm looking for. I need to clarify the question. In this case $f(x,y)$ is not Riemann integrable at a single point (measure zero) so $\int_Bf(x,y)dy$can be arbitrarily defined at that point and is still integrable almost everywhere and $\int_Q f = \int_a\int_b f(x,y) dy dx$. I gave you an upvote anyway. $\endgroup$
    – AlRacoon
    Commented Jan 24, 2018 at 16:53
  • $\begingroup$ Also in this example, $f$ is nonnegative and absolutely integrable -- which I mentioned above is not what I was interested in. $\endgroup$
    – AlRacoon
    Commented Jan 24, 2018 at 16:57

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