Let $f: \Bbb R \to \Bbb R$ be Borel-measurable. $(\mathrm{A})$ Show that $g:\Bbb R ^2 \to \Bbb R, g(x,t)=f(x+t) $ is Lebesgue-measurable. $(\mathrm{B})$ Assume $g \ne 0$ (meaning $g$ is not a.e. zero). Show that $g$ is not Lebesgue integrable.
Part $\mathrm{B}$ is rather simple, as we can easily create a set of full measure at which $g$ is nonzero. But I couldn't find a clean way to use the weaker notion of $f$ being Borel-measurable in part $\mathrm{A}$. Any help?
Define $h: \mathbb R^{2} \to \mathbb R$ by $h(x,t)=x+t$. Then $h$ is Borel measurable because it is continuous. $g=f\circ h$, so $g$ is Borel measurable. This implies that it is also Lebesgue measurable.
The uncertainty about (A) most likely stems from a confusion about definitions stemming from the unfortunate naming of "Lebesgue-measurable" functions.
Namely, it is usually the case (and for the result of this question to hold, must be the case) that a function $f$ is called "Lebesgue-measurable" if the preimage $f^{-1}(B)$ of any Borel set $B$ is in the Lebesgue $\sigma$-algebra. Reasons for this definition are discussed here: Why this definition for Lebesgue measurable functions?
This is weaker than the preimages $f^{-1}(L)$ of Lebesgue sets $L$ having to be Lebesgue, which is what the name "Lebesgue-measurable" incorrectly suggests.
It follows from this definition that all Borel-measurable functions are also Lebesgue-measurable. In particular, $g$ is a composition of two Borel-measurable functions ($f$ and addition), which makes it Borel- and so Lebesgue-measurable.
PS. It's not enough for $g$ to be positive on a full measure set for it to not be integrable - think of $h(x, y) = \exp\{-(x^2 + y^2)\}$, which in integrable despite being everywhere positive.
A different approach would be to find an $A$ such that $\lambda(A) > 0$ and $f(x) > \epsilon$ for all $x \in A$ and some $\epsilon > 0$ and show that $g(x, t) > \epsilon$ on the set $$ B = \biggl\{ \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} : \begin{pmatrix} x \\ y \end{pmatrix} \in A \times \mathbf{R} \biggr\}, $$ which is a shear of $A \times \mathbf{R}$ and so has infinite Lebesgue measure.
