2.4

It suffices to consider $f$ to be a non-negative $L^1$ function. It's not at first apparent that $g(x)$ is an $L^1$ function. First, define the set: $$T(b) = \lbrace (x,t) \in \mathbb{R} \times \mathbb{R} \hspace{0.25cm} \big|\hspace{0.25cm} 0 < x \leq t, \hspace{0.25cm} x \leq t \leq b \rbrace$$ (It's just the triangle formed by taking the square $[0,b] \times [0,b]$, bisecting it with the line $x = t$, and taking the top half.) Next, define the function: $$F(x,t) = \frac{f(t)}{t} \chi_{T(b)}$$ Certainly, $F$ is measurable, since $\chi_{T(b)}$, $f$, and $\frac{1}{t}$ are measurable functions. Next, observe that:

$$\int_{[0,b]} g(x) dx = \int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{\mathbb{R}} \int_{\mathbb{R}} \frac{f(t)}{t} \chi_{T(b)} dt dx$$ From here, Tonelli's Theorem gives, that since $F$ is measurable: $$\int_{[0,b]} \int_{[x,b]} \frac{f(t)}{t}dtdx = \int_{[0,b]} \int_{[0,t]} \frac{f(t)}{t} dx dt = \int_{[0,b]} f(t) dt$$ Thus, since $f \in L^1$, we have: $$\int_{[0,b]} g(x) dx = \int_{[0,b]} f(t) dt < \infty \hspace{0.25cm} \Rightarrow g \in L^1$$ ...as desired.