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Let $f:[0,1]\times [0,2\pi]\to \mathbb R$ be an element of $L_1[0,1]$ of the first parameter and be a continuous function of the second parameter. Can we say that $f\in L_1([0,1]\times [0,2\pi])$?

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Messing around a bit, I believe that $$f(x,y) = \begin{cases} \displaystyle \frac{y}{y^2 + e^{-1/x}}, & x \ne 0 \\ 0, & x =0 \end{cases}$$ is a counterexample. It is clearly continuous in $y$. For any fixed $y \ne 0$, $f(\cdot, y)$ is a bounded function, hence in $L^1$, and $f(\cdot, 0) = 0$. Since $f$ is nonnegative, by Tonelli's theorem we may compute $\iint_{[0,1] \times [0,2\pi]} f$ as $$\begin{align*} \int_0^1 \int_0^{2\pi} f(x,y)\,dy\,dx &= \frac{1}{2} \int_0^1 \left(\ln(e^{-1/x} + (2\pi)^2) - \ln(e^{-1/x}))\right)\,dx \\ &= \frac{1}{2} \left(\int_0^1 \ln(e^{-1/x} + (2\pi)^2)\,dx + \int_0^1 \frac{1}{x}\,dx\right). \end{align*}$$ The first integral is finite since the integrand is bounded, and the second integral is infinite.

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