Integrating continuous function of two variables by one of the variables, do I get continuous function?

Question

Let $f(x,y)$ be a function continuous in $x$ such that $g(x) = \int_a^b f(x,y) dy$ exists for every $x$. Is $g(x)$ necessarily continuous? I am especially interested in the Riemann and Lebesgue integrals.

Answer 1

Let us take any sequence $(x_n)_n$ converging towards $x$. Consider $f_n(y)=f(x_n,y)$. By continuity of $f$, each $f_n$ is Lebesgue integrable. Furthermore, $f_n\to f$ for $n\to \infty$. Now, $$g(x)=\int f(x,y)dy=\int \lim\limits_n f_n(y)dy=\lim\limits_n\int f_n(y)dy=\lim\limits_n g(x_n),$$ where we used the Dominated Convergence Theorem to interchange limit and integration. But what we just wrote is the sequential criterion for continuity.

Answer 2

There is a counterexample $$f(x,y)=\left\{\begin{matrix} 0&y=0\\ \frac{1}{y^2}\max\left\{y-|x-y|,0\right\}&y\neq0 \end{matrix}\right.$$ $f(x,y)$ is countinuous expect for $(0,0)$. But for $0<x<1$ we have $$\begin{align*} g(x)&:=\int_0^1f(x,y)dy \geq\int_{3x/4}^x\frac{1}{y^2}\left(y-|x-y|\right)dy\\ &\geq\frac{1}{x^2}\cdot \frac{x}{4}\cdot(\frac{3x}{4}-\frac{x}{4})=\frac{1}{8} \end{align*}$$ Since $g(0)$ is obviously $0$, $g(x)$ is discontinuous at $x=0$