1. ## Integral proof

hello could someone help me with this problem i have an ans but i dont think its worked fully.thanks
Prove that if f and g are integrable functions on [a,b] then f+g is integrable and

2. Originally Posted by matty888
hello could someone help me with this problem i have an ans but i dont think its worked fully.thanks
Prove that if f and g are integrable functions on [a,b] then f+g is integrable and

Let $P(x_0, x_1, ..., x_n)$ be a partition of $[a,b]$ with $I_k=[x_{k-1},x_k]$.

recall that $sup(f+g) \leq sup \, f + sup \, g$ and $inf(f+g) \geq inf \, f + inf \, g$ over the interval.

now, for any partition $P$,
$L(P,f) + L(P,g) \leq L(P, f+g) = \sum_{k=1} ^n inf(f+g)(x_k-x_{k-1})$ ---- (1)

$U(P,f) + U(P,g) \geq U(P, f+g) = \sum_{k=1} ^n sup(f+g)(x_k-x_{k-1})$ ---- (2)

So, let $\varepsilon > 0$. since $f$ and $g$ are integrable on $[a,b]$,

$\exists \, P_{f,\varepsilon}, \, P_{g,\varepsilon}$ such that

$U(P_{f,\varepsilon},f) - L(P_{f,\varepsilon},f) < \frac{\varepsilon}{2}$ --- (3)

$U(P_{g,\varepsilon},f) - L(P_{g,\varepsilon},f) < \frac{\varepsilon}{2}$ --- (4)

so, take $P_{\varepsilon} = P_{f,\varepsilon} \cup P_{g,\varepsilon}$ which is a refinement of $P_{f,\varepsilon}, \, P_{g,\varepsilon}$..

$U(P_{\varepsilon}, f+g) \leq U(P_{\varepsilon}, f) + U(P_{\varepsilon}, g)$ by (2)

$\leq U(P_{f,\varepsilon},f) + U(P_{g,\varepsilon},g)$ (from theorem or lemma)

$\leq L(P_{f,\varepsilon},f) + L(P_{g,\varepsilon},g) + \varepsilon$ by (3) and (4)

$\leq L(P_{\varepsilon},f) + L(P_{\varepsilon},g) + \varepsilon$ (from theorem or lemma)

$\leq L(P_{\varepsilon},f+g) + \varepsilon$ by (1)

and so, $U(P_{\varepsilon},f+g) - L(P_{\varepsilon},f+g) \leq \varepsilon$ which means $f+g$ is integrable on $[a,b]$..

for the second part, you can do it by showing one is greater or equal than the other and vice versa..