"Half"-linear real function Let $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+y)=f(x)+f(y)$, I want to show that:
(a) If $f$ bounded in an interval $\Rightarrow$ $f$ continuous.
(b) If $f$ bounded in a set with positive Lebesgue measure $\Rightarrow$ $f$ continuous.
(c) If $f$ Lebesgue-measurable $\Rightarrow$ $f$ bounded around the origin
Edit:
I forgot to put what I have done so far, which isn't much sadly. I showed that $(a)\Rightarrow (b)$, since that set has to contain a compact set with positive measure, i.e. a closed interval.
Edit 2:
As pointed out in the comments my previous approach was wrong, I think I corrected it: Since $\lambda(A)>0\Rightarrow A-A$ contains a neighbourhood  of the origin, where $f$ is bounded, therefore $(a)$ implies $(b)$.
 A: We use the following

Lemma: If $f \colon \mathbb{R} \to \mathbb{R}$ is an additive function (that is, $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{R}$), then $f$ is continuous on $\mathbb{R}$ if and only if there is an $x_0\in \mathbb{R}$ such that $f$ is continuous at $x_0$.

The one direction is immediate, and for the other we note that $f$ is continuous at $x_0$ if and only if $f$ is continuous at $0$: $f$ is continuous at $x_0$ if and only if for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\lvert y-x\rvert < \delta \implies \lvert f(y) - f(x)\rvert < \varepsilon$. That is, $\lvert h\rvert < \delta \implies \lvert f(x+h) - f(x)\rvert < \varepsilon$, and since $f(x+h) - f(x) = f(h)$ and $f(0) = 0$, this is equivalent to $\lvert h-0\rvert < \delta \implies \lvert f(h) - f(0)\rvert < \varepsilon$.
Next we show that if an additive $f$ is bounded on some neighbourhood of $0$, then it is continuous at $0$ (and by the lemma, on all of $\mathbb{R}$).
So suppose there is a $c > 0$ and an $M \in \mathbb{R}$ with $\lvert x\rvert < c \implies \lvert f(x)\rvert \leqslant M$. By the additivity, we have $f(k\cdot y) = k\cdot f(y)$ for all $k\in \mathbb{Z}$ and $y\in \mathbb{R}$. Given $\varepsilon > 0$, choose $n\in \mathbb{N}\setminus \{0\}$ so large that $\frac{M}{n} < \varepsilon$. Then for $\lvert x\rvert < \frac{c}{n}$ we have
$$\lvert f(x)\rvert = \biggl\lvert \frac{n}{n}f(x)\biggr\rvert = \biggl\lvert \frac{1}{n} f(nx)\biggr\rvert = \frac{1}{n}\lvert f(nx)\rvert \leqslant \frac{1}{n}M < \varepsilon.$$
Thus $f$ is continuous at $0$.
(a) If $f$ is bounded on a nondegenerate interval, say $\lvert f(x)\rvert \leqslant M$ for $x \in (a,b)$ with $a < b$, then $f$ is bounded on a neighbourhood of $0$: Let $\mu = \frac{a+b}{2}$ and $c = \frac{b-a}{2}$. For $\lvert x\rvert < c$, we have $a < x+\mu < b$ and therefore $\lvert f(x+\mu)\rvert \leqslant M$. Hence
$$\lvert f(x)\rvert = \lvert f(x+\mu) - f(\mu)\rvert \leqslant \lvert f(x+\mu)\rvert + \lvert f(\mu)\rvert \leqslant 2M,$$
so $f$ is bounded on $(-c,c)$, and by the above, $f$ is continuous.
(b) If $f$ is bounded on a set with positive Lebesgue measure, say $\lvert f(x)\rvert \leqslant M$ for $x\in A$ with $\lambda(A) > 0$, then $f$ is bounded on some nondegenerate interval, and by (a), it follows that $f$ is continuous. To see that $f$ is bounded on some nondegenerate interval, note that since $\lambda(A) > 0$, the set $A + A$ contains a nondegenerate interval, and for $x = a_1 + a_2 \in A + A$ we have
$$\lvert f(x)\rvert = \lvert f(a_1) + f(a_2)\rvert \leqslant \lvert f(a_1)\rvert + \lvert f(a_2)\rvert \leqslant 2M.$$
(c) If $f$ is Lebesgue measurable, then $f$ is bounded on a set of positive Lebesgue measure, and hence - by (b) - continuous (and therefore bounded on some neighbourhood of $0$).
Since $f$ is measurable, the sets
$$A_n := \{ x \in \mathbb{R} : \lvert f(x)\rvert \leqslant n\}$$
are Lebesgue measurable, and we have $A_n \subset A_{n+1}$ for all $n\in \mathbb{N}$ and
$$\mathbb{R} = \bigcup_{n = 0}^{\infty} A_n.$$
By the continuity from below of measures, we have
$$\lim_{n\to \infty} \lambda(A_n) = \lambda(\mathbb{R}) = +\infty,$$
in particular $\lambda(A_n) > 0$ for all sufficiently large $n$.
