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I need help with the following task (dealing with the Riemann-Integration), I am doing as an exercise for myself right now.

Let $f:[a,b]\to\mathbb{R}$ be integrable and $\alpha \in \mathbb{R}$. Then $\alpha \cdot f$ is integrable with $\int_a^b \alpha f=\alpha \int_a^b f$

The thing is we already proved this task: Let $f,g:[a,b]\to\mathbb{R}$ be integrable. Then $f+g$ is integrable with $\int_a^b (f+g)= \int_a^b f + \int_a^b g$. Since we were told, that we can prove the task above in a similar way, I wanted to do this that way. Here is the proof for the already proved task:

Since f,g are integrable, they are bounded, thus f+g is bounded. Let $Z$={$x_0,...,x_n$} be a partition of [a,b]. Since $\underset{x\in [x_{j-1},x_j]} \sup (f(x)+g(x))\leq \underset{x\in [x_{j-1},x_j]} \sup f(x) +\underset{x\in [x_{j-1},x_j]} \sup g(x)$, $j=1,...,n$ we have $$U_{f+g}(Z)\leq U_f(Z)+U_g(Z)$$ (with $U_f(Z)=\displaystyle\sum_{j=1}^n$ $\underset{x\in [x_{j-1},x_j]} \sup f(x) (x_j-x_{j-1}))$. Therefore we get $$\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z)+U_g(Z)$$

($\underset{Z} \inf U_{f+g}(Z)$ is the infimum of the set of all upper Darboux sums of $f+g$ regarding Z)

Let $Z_1$ and $Z_2$ be any two partitions of [a,b]. For $Z= Z_1 \cup Z_2$, we already proved that $$U_f(Z_1)\geq U_f(Z),$$ $$U_g(Z_2)\geq U_g (Z)$$ So therefore we have $$\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z)+U_g(Z)\implies\underset{Z} \inf U_{f+g}(Z)\leq U_f(Z_1)+U_g(Z_2)$$ which implies that $$\underset{Z} \inf U_{f+g}(Z)\leq \underset{Z_1} \inf U_f(Z_1)+ \underset{Z_2} \inf U_g(Z_2)$$ $$\underset{Z} \inf U_{f+g}(Z)\leq \underset{Z} \inf U_f(Z)+ \underset{Z} \inf U_g(Z)$$

With help of the same argument we get $$\underset{Z}\sup L_{f+g}(Z)\geq \underset {Z} \sup L_f(Z)+ \underset {Z}\sup L_g(Z)$$ with $L_f(Z)=\sum_{j=1}^n \underset {x\in[x_{j-1},x_j]} \inf f(x) (x_j-x_{j-1})$.

Thus we get $$\underset{Z} \inf U_{f+g}(Z)\leq \underset{Z} \inf U_f(Z)+ \underset{Z} \inf U_g(Z)$$ $$= \underset{Z}\sup L_f(Z)+\underset{Z}\sup L_g(Z)$$ (since f,g integrable over [a,b]) $$\leq \underset{Z}\sup L_{f+g}(Z)$$

Since also $$\underset{Z} \inf U_{f+g}(Z)\geq \underset{Z}\sup L_{f+g}(Z)$$ we have $$\underset{Z} \inf U_{f+g}(Z)= \underset{Z}\sup L_{f+g}(Z)$$ and therefore f+g is integrable over [a,b]. Furthermore we get $$\underbrace{\underset{Z}\sup L_{f+g}(Z)}_{=\int_a^b (f+g)}=\underbrace{\underset{Z}\sup L_f(Z)}_{=\int_a^b f}+\underbrace{\underset{Z}\sup L_g(Z)}_{=\int_a^b g}$$

So how are we gonna do this for $\int_a^b \alpha f=\alpha \int_a^b f$ now? My idea was to start like in the proof above. Since $f$ is bounded and $\alpha$ is a constant, $\alpha \cdot f$ is bounded too. But we can't say that $\underset{x\in [x_{j-1},x_j]} \sup (f(x)\alpha)\leq \underset{x\in [x_{j-1},x_j]} \sup f(x) \cdot \alpha,$ $j=1,...,n$, since $\alpha$ can be negative too,... thats why I don't really see how to start.

Is there anyone who could give me an advice? I would be very grateful.

Analysis_Mark
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    You should mention that it is Riemann-integration. – Jean Marie Apr 15 '22 at 11:40
  • Just consider cases for $\alpha$. For $\alpha>0$, your last consideration is true. Then just use the fact that if $f$ is intergradable then $-f$ is also. – dmtri Apr 15 '22 at 11:47
  • Sorry it was interupted....And $\int_a^b -g=-\int_a^b g$ – dmtri Apr 15 '22 at 11:56
  • @dmtri Okay, for $\alpha > 0$, we then get $\underset {Z}\inf U_{\alpha f}(Z)\leq \alpha \cdot U_f(Z)$. In the proved task, we take any two partitions $Z_1$, $Z_2$ and say $Z=Z_1\cup Z_2$, so that we can say $U_f(Z_1)\geq U_f(Z)$ and $U_g(Z_2)\geq U_g(Z)$. How should I do this here. Just saying $U_f(Z_1)\geq U_f(Z)$ such that we then have $\underset {Z}\inf U_{f\alpha}\leq \alpha \cdot O_f(Z_1)$, without really using $Z_2$? – Analysis_Mark Apr 15 '22 at 12:10
  • Why not $\sup\alpha f=\alpha \sup f$, If $\alpha>0$? – dmtri Apr 15 '22 at 15:55

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