# Product of two integrable function is integrable with respect to product measure

I know this question has been answered before, but it has not answered my confusion.

The question is like this:

$f$ and $g$ are integrable, show that $h(x,y)=f(x)g(y)$ is integrable with respect to product measure then $\int f(x)g(y) d(\mu\times \nu)=\int f(x)(\int g(y)d\nu)d\mu$. And a hint for this problem is to use Fubini, and $\mu,v$ is not necessarily sigma finite.

I am thinking that in order to use Fubini's theorem, $h$ must be integrable. I can show that this is true by building up from characteristic functions, but it will get $\int f(x)g(y) d(\mu\times \nu)=\int f(x)(\int g(y)d\nu)d\mu$ along the process. So this the hint true or ???

The other answer and the comments therein do not actually answer this question. Also Fubini's Theorem does not assume $$\sigma$$-finiteness, but rather completeness, of the measure spaces. Let's say we're working with the measure spaces $$(X, \mathfrak{A}, \mu)$$ and $$(Y, \mathfrak{B}, \nu)$$. The hypothesis, then, is that $$\mu$$ and $$\nu$$ are complete measures (I'd assume), and that $$f \in L^1(X, \mathfrak{A}, \mu)$$ and $$g \in L^1(Y, \mathfrak{B}, \nu)$$.

We want to show that $$h(x, y) := f(x)g(y)$$ is $$\mu \times \nu$$-integrable, and that

$$\begin{equation*} \int h(x, y) d(\mu \times \nu)(x, y) = \int f(x) d\mu(x) \cdot \int g(y) d\nu(y) \end{equation*}$$

This can done by painful bootstrapping (though there is probably a simpler procedure):

1. Suppose $$f(x)$$ and $$g(y)$$ are simple functions, say, $$f(x) = \sum\limits_{m=1}^{M} a_n \chi_{A_m}(x)$$ and $$g(y) = \sum\limits_{n=1}^{N} b_n \chi_{B_n}(y)$$ in canonical form (i.e., each $$A_m$$ is the preimage of $$\{ a_m \}$$, and each $$B_n$$ is the preimage of $$\{ b_n \}$$). Then

\begin{align*} \int f(x)g(y) d(\mu \times \nu)(x, y) &= \int \left( \sum\limits_{m=1}^{M} a_m \chi_{A_m}(x) \right) \cdot \left( \sum\limits_{n=1}^{N} b_n \chi_{B_n}(y) \right) d(\mu \times \nu) (x, y) \\ &= \sum\limits_{m=1}^{N} \sum\limits_{n=1}^{N} a_m \cdot b_n \cdot \int \left( \chi_{A_m}(x) \chi_{B_n}(y) \right) d(\mu \times \nu) (x, y) \end{align*}

By the definition of a simple function (in canonical form), $$A_m$$ and $$B_n$$ are measurable for all $$1 \le n \le N$$ and $$1 \le m \le M$$. So,

\begin{align*} \int\chi_{A_m}(x) \chi_{B_n}(y) d(\mu \times \nu)(x, y) &= \int \chi_{A_m \times B_n} (x, y) d(\mu \times \nu)(x, y) \\ &= (\mu \times \nu) (A_m \times B_n) \\ &= \mu(A_m) \nu(B_n) \end{align*}

which must be finite since $$A_m$$ and $$B_n$$ are measurable. $$\int fg d(\mu \times \nu)$$ is, as shown above, a linear combination of these finite values, so $$fg$$ is $$\mu \times \nu$$-integrable. Also you can rearrange the sum:

\begin{align*} \sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} a_m \cdot b_n \cdot \int \left( \chi_{A_m}(x) \chi_{B_n}(y) \right) d(\mu \times \nu) (x, y) &= \sum\limits_{m=1}^{M} \sum\limits_{n=1}^{N} a_m \cdot b_n \cdot \mu(A_m) \nu(B_n) \\ &= \left( \sum_{m=1}^{M} a_m \mu(A_m) \right) \cdot \left( \sum_{n=1}^{N} b_n \nu(B_n) \right) \\ &= \left( \sum_{m=1}^{M} a_m \int \chi_{A_m} (x) d\mu(x) \right) \cdot \left( \sum_{n=1}^{N} b_n \int \chi_{B_n}(y) d\nu(y) \right) \\ &= \left( \int \left[ \sum\limits_{m=1}^{M} a_m \chi_{A_m}(x) \right] \right) \cdot \left( \int \left[ \sum\limits_{n=1}^{N} b_n \chi_{A_n}(x) \right] \right) \\ &= \left( \int f(x) d\mu(x) \right) \cdot \left( \int g(y) d\nu(y) \right) \end{align*}

So, long story short, $$\int fg d(\mu \times \nu) = \int f d\mu \cdot \int g d\nu$$.

1. Suppose $$f(x)$$ is $$\mu$$-integrable and $$g(x)$$ is $$\nu$$-integrable, and both are nonnegative. Then there exist two increasing sequences of simple functions, $$\{ F_m \}_{m=1}^{\infty}$$ and $$\{ G_n \}_{n=1}^{\infty}$$, such that $$F_m \to f$$ almost everywhere [$$\mu$$] and $$G_n \to g$$ almost everywhere [$$\nu$$]. (I honestly don't remember how to prove that.) Therefore, $$F_n(x) G_n(y) \to f(x)g(y)$$ almost everywhere [$$\mu \times \nu$$]. By Case 1 and the Monotone Convergence Theorem, $$f(x)g(y) \in L^1 (\mu \times \nu)$$ and

\begin{align*} \int f(x)g(y) d\mu(\mu \times \nu)(x, y) &= \lim\limits_{n \to \infty} \int F_n(x) G_n(y) d(\mu \times \nu)(x, y) \\ &= \lim\limits_{n \to \infty} \left( \int F_n(x) d\mu(x) \cdot \int G_n(y) d\nu(y) \right) \\ &= \left( \lim\limits_{n \to \infty} \int F_n(x) d\mu(x) \right) \cdot \left( \lim\limits_{n \to \infty} \int G_n(y) d\nu(y) \right) \\ &= \left( \int f(x) d\mu(x) \right) \cdot \left( \int g(x) d\nu(x) \right) \end{align*}

(Apply the MCT to $$f$$, $$g$$ and $$fg$$.)

1. Finally, let's return to the original hypothesis that $$f$$ and $$g$$ are merely integrable. We can now split $$h(x, y) = f(x) g(y)$$ into its negative and positive parts: $$h(x, y) = h^+(x, y) - h^-(x, y)$$. We can rewrite these as

\begin{align*} h^+(x, y) = f^+(x) g^+(y) + f^-(x) g^-(y) \\ h^-(x, y) = f^+(x) g^-(y) + f^-(x) g^+(y) \end{align*}

Case 2 applies to the functions $$f^+ f^+$$, $$f^+ g^-$$, $$f^- g^+$$ and $$f^- g^-$$. So now (omitting the domain variables for some hope of readability):

\begin{align*} \int h \;d(\mu \times \nu) &= \int h^+ d(\mu \times \nu) - \int h^- d(\mu \times \nu) \\ &= \int \left(f^+ g^+ + f^- g^-\right)d(\mu \times \nu) - \int \left(f^+ g^- + f^- g^+ \right) d(\mu \times \nu) \\ &= \int f^+ d\mu \int g^+ d\nu + \int f^- d\mu \int g^- d\nu - \int f^+ d\mu \int g^- d\mu - \int f^- d\mu \int g^+ d\nu \\ &= \left(\int f^+ d\mu - \int f^- d\mu \right) \cdot \left( \int g^+ d\nu - \int g^- d\nu \right) \\ &= \int f d\mu \int g d\nu \end{align*}

• On the contrary, Fubini's theorem does assume σ-finiteness, not completeness. As a counterexample, take $X = Y = [0, 1]$ with $\mu$ the Lebesgue measure, and $\nu$ the counting measure (on all subsets $E \subseteq [0, 1]$.) The product measure isn't unique, so take the minimal one: the measure of a measurable subset of $[0, 1]^2$ is the sum of $\mu$-measures of its slices. Let $A$ be the diagonal $\{(x, x) : x \in [0, 1]\}$. Then $(\mu \times \nu)(A) = 0$, so the indicator $1_A$ is integrable, but one of the iterated integrals evaluates to $1$. Nov 21, 2022 at 22:15
• Actually, merely $\Sigma$-finiteness of both measures is sufficient (a measure is called $\Sigma$-finite if it's the sum of a sequence of finite measures.) Completeness is not required (there do exist variants of Fubini for compete measures which are modified slightly, but these variants do require $\sigma$-finiteness, too.) Actually, it's sufficient for merely the measure of the "inner" iterated integral to be σ-finite, provided the other measure is complete (or satisfies one of a couple other technical conditions). See Fremlin's Measure Theory, Volume 2, result 252B. Nov 21, 2022 at 22:38

Fubini's theorem states that if one of the following integrals

$$\int \left(\int |h(x,y)| \, d\mu(x)\right) d\nu(y) \quad \int \left(\int |h(x,y)| \, d\nu(y) \right) d\mu(x) \quad \iint |h(x,y)| \, d(\mu \otimes \nu)(x,y)$$

is finite, then all of these integrals are finite and

$$\int \left( \int h(x,y) \, d\mu(x) \right) d\mu(y) = \int \left( \int h(x,y) \, d\nu(y) \right) d\mu(x) = \iint h(x,y) \, d(\mu \otimes \nu)(x,y).$$

In particular, if $h(x,y) = f(x) \cdot g(y)$ with integrable $f$, $g$, then by the linearity of the integral

$$\int |h(x,y)| \, d\mu(x) = |g(y)| \left( \int |f| \, d\mu \right)$$

and therefore

$$\int \left(\int |h(x,y)| \, d\mu(x) \right) d\nu(y) = \left( \int |f| \, d\mu \right) \left( \int |g| \, d\nu \right) < \infty.$$

This means that we may apply Fubini's theorem to conclude that we can interchange the integrals. Note that this holds only true if $\mu$ and $\nu$ are $\sigma$-finite measures.

• Don't we need the assumption that h is integrable with the product measure? Dec 19, 2015 at 19:55
• No, we don't need this. If one of the integrals in the second line (of my answer) is finite, then all these integrals are finite; in particular, $h$ is integrable with respect to the product measure
– saz
Dec 19, 2015 at 20:12
• Thanks ! Can you give a link to a reference to this ? since I am using Royden's real analysis, the Fubini theorem there has the assumption that it is intgetrable w.r.t. the product measure, and the Tonelli's theorem has the assumption that $\mu, v$ are sigma-finite Dec 19, 2015 at 20:19
• @starry1990 Since you didn't mention it in your question, I didn't know that you are interested in the case where the measures are not $\sigma$-finite. If the measures are not $\sigma$-finite, then we really need that $h$ is integrable wrt the product measure, i.e. $$\iint |h(x,y)| \, d(\mu \otimes \nu)(x,y)<\infty.$$ see also this question: math.stackexchange.com/questions/887491/…
– saz
Dec 19, 2015 at 20:35
• Then we can only do this by starting from characteristic function? Dec 19, 2015 at 20:36