Let $f\in L^1(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$ be integrable and bounded.
Question: Does $x\mapsto \int_{\mathbb{R}} |f(x,y)| \mathrm{d}y$ define a bounded function? In other words, is the following true? $$\sup_{x\in\mathbb{R}} \int_{\mathbb{R}} |f(x,y)| \mathrm{d}y<\infty.$$
That is, if we integrate $f$ over one parameter, is the resulting function bounded? There are a few special cases in which this is definitely true, for example, if $f(x,y)=g(x)h(y)$ for some functions $g,h$. Then $f\in L^1(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$ implies $g,h\in L^1(\mathbb{R})\cap L^\infty(\mathbb{R})$ and $$\sup_{x\in\mathbb{R}} \int_{\mathbb{R}} |f(x,y)| \mathrm{d} y = \sup_{x\in\mathbb{R}}|g(x)| \int_{\mathbb{R}} |h(y)| \mathrm{d} y <\infty.$$
Moreover, if $f$ is compactly supported in $x$ (i.e. $f(x,y)=0$ if $x\notin K$, $K\subset \mathbb{R}$ compact), then $$\sup_{x\in\mathbb{R}} \int_{\mathbb{R}} |f(x,y)| \mathrm{d}y \leq |K| \sup_{x,y\in\mathbb{R}} |f(x,y)| < \infty.$$ However, I cannot prove it for general $f$. Also, I did not find a counterexample.
If you care about every single line, then just having $f(x, y) = \mathbb I_{x = 0}$ is enough. If you want an example where even essential supremum is infinite, construction is a bit more complicated.
Let $$f(x, y) = \begin{cases} 1, x \in [2^{-k}, 2^{-k + 1}), y \in [0, k] \\ 0\end{cases}$$
Then $f$ is bounded by $1$, integral of it is $\sum\limits_k k \cdot 2^{-k} < \infty$. But $\int_\mathbb R f(2^{-k}, y)\, dy = k$, which is not bounded.