Improper Riemann integrabilty of Lebesgue integrable function.

Question

Suppose $f(x, y)$ is a continuous and Lebesgue integrable function on $[a; b) \times [c; d)$, $b, d \in \mathbb{R} \cup \{+\infty\}$. Riemann integral $\int_a^b(\int_c^d|f(x, y)|dy)dx$ converges.

Is it true that $I(x) = \int_{[c; d]}f(x, y)d\mu(y) = \int_c^df(x,y)dy$ is a continuous function? Or, in other words, is it true that $\int_{[a; b) \times [c; d)}f(x,y)d\mu(x, y) = \int_{[a; b]}(\int_{[c; d]}f(x, y)d\mu(y))d\mu(x) = \int_{[a; b]}(\int_c^df(x,y)dy)d\mu(x) =^? \int_a^b(\int_c^df(x, y)dy)dx?$

Answer 1

1. First I would like to touch on the the continuity of $F(x):=\int^d_cf(x,y)\,\nu(dy)$ under the assumption that $f\in C([a,b)\times[c,d))\cap L_1([a,b)\times[c,d),\mu\otimes\nu))$. Problems may arise when the intervals are infinite and the measures where integration is being carried are infinite.

Suppose $d=\infty$ and $\nu(dy)$ is Lebesgue measure.

• If $|f(x,y)-f(x',y)|\leq g(y)$ for some $\nu$-integrable function $g$, continuity of $F$ follows by dominated convergence for $$ |F(x)-F(x_0)|\leq\int^\infty_c|f(x,y)-f(x_0,y)|\,d\nu(y). $$
• Without additional assumptions on $f$ as the one above, continuity may fail as the next counterexample shows. Let $$\lambda(x)=\frac{1}{(1-x)^{1/2}}\mathbb{1}_{[0,1)}+\frac{1}{(x-1)^2}\mathbb{1}_{(1,\infty)},$$ and define $$ f(x,y)=\left\{\begin{matrix} |x-1|^{1/2}e^{-y/\lambda(x)}& x\neq1\\ 0 &x=1 \end{matrix} \right. $$ $f$ is continuous in $[0,\infty)\times[0,\infty)$ (not too difficult to check) $$\int^\infty_0\int^\infty_0f(x,y)\,dy\,dx=\int^\infty_0|x-1|^{1/2}\lambda(x)\,dx<\infty$$ This, along with $f>0$, implies that $f$ in Lebesgue integrable in $\mathbb{R}^2$. On the other hand, $$ F(x):=\int^\infty_0f(x,y)\,dy=\left\{ \begin{matrix} |1-x|^{1/2}\lambda(x)& x\neq1\\ 0 & x=1 \end{matrix} \right. $$ is discontinuous at $x=1$

2. If the intevals of integration are finite and $f\in C([a,b]\times[c,d])$ then continuity of $F(x)=\int^d_cf(x,y)\,\mu(dy)$ is straight forward from dominated convergence.

3. The indetity $$\int_{[a,b)\times[c,d)}f=\int_{[a,b)}\int_{[c,d)}f$$ where $f$ is Lebesgue integrable has nothing to do with the continuity (or lack thereof) of $f$. This is one of the conclusions of Fubini-Tonelli's theorem, that is, if the measures $\mu$ and $\nu$ on $X$ and $Y$ respectively are $\sigma$-finite, and $$\int_X\Big(\int_Y |f(x,y)|\,\nu(dy)\big)\,\mu(dx<\infty,$$ then $f$ is $\mu\otimes\nu$-integrable and $\int_{X\times Y}f d(\mu\times\nu)=\int_X\Big(\int_Yfd\nu\Big)\,d\mu=\int_Y\Big(\int_Xf\,d\mu\Big)\,d\nu$.